Practice · Place consecutive whole numbers to sum decimals · Sensim Math

Variant 1 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $4$ and less than $5$ , find $100$ times $A.BC$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 4 and 5. Find 100 times A.BC.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
4 < A.BC + C.DE < 5.

Unknowns

The value of 100 x A.BC.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 4 and 5.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 4 and 5

#6 Guess and Check 5.NBT.A.3

We need the sum near 4, so 2A+2 should be about 4. Test A=1: the digits are 1, 2, 3, 4, 5, so the decimals are A.BC = 1.23 and C.DE = 3.45.

2A+2 \approx 4 \Rightarrow A=1 \Rightarrow A.BC=1.23,\ C.DE=3.45

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 1.23 + 3.45 = 4.68, which is greater than 4 and less than 5. So A=1 fits and A.BC = 1.23.

1.23+3.45=4.68,\quad 4<4.68<5

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 1.23 into 123.

1.23 \times 100 = 123

Times 100 moves the point two spots, a place-value shift.

Answer: 123

4 · Reviewdoes it hold up?

The digits 1, 2, 3, 4, 5 are consecutive and single-digit. The sum 4.68 is indeed between 4 and 5. Multiplying 1.23 by 100 gives 123, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (4, 5); only A=1 survives, giving 100 x 1.23 = 123.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 1.23 + 3.45 and multiplying 1.23 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 1.23 and 3.45 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 2 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $6$ and less than $7$ , find $100$ times $A.BC$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 6 and 7. Find 100 times A.BC.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
6 < A.BC + C.DE < 7.

Unknowns

The value of 100 x A.BC.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 6 and 7.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 6 and 7

#6 Guess and Check 5.NBT.A.3

We need the sum near 6, so 2A+2 should be about 6. Test A=2: the digits are 2, 3, 4, 5, 6, so the decimals are A.BC = 2.34 and C.DE = 4.56.

2A+2 \approx 6 \Rightarrow A=2 \Rightarrow A.BC=2.34,\ C.DE=4.56

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 2.34 + 4.56 = 6.90, which is greater than 6 and less than 7. So A=2 fits and A.BC = 2.34.

2.34+4.56=6.90,\quad 6<6.90<7

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 2.34 into 234.

2.34 \times 100 = 234

Times 100 moves the point two spots, a place-value shift.

Answer: 234

4 · Reviewdoes it hold up?

The digits 2, 3, 4, 5, 6 are consecutive and single-digit. The sum 6.90 is indeed between 6 and 7. Multiplying 2.34 by 100 gives 234, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (6, 7); only A=2 survives, giving 100 x 2.34 = 234.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 2.34 + 4.56 and multiplying 2.34 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 2.34 and 4.56 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 3 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $6$ and less than $7$ , find $100$ times $C.DE$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 6 and 7. Find 100 times C.DE.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
6 < A.BC + C.DE < 7.

Unknowns

The value of 100 x C.DE.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 6 and 7.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 6 and 7

#6 Guess and Check 5.NBT.A.3

We need the sum near 6, so 2A+2 should be about 6. Test A=2: the digits are 2, 3, 4, 5, 6, so the decimals are A.BC = 2.34 and C.DE = 4.56.

2A+2 \approx 6 \Rightarrow A=2 \Rightarrow A.BC=2.34,\ C.DE=4.56

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 2.34 + 4.56 = 6.90, which is greater than 6 and less than 7. So A=2 fits and C.DE = 4.56.

2.34+4.56=6.90,\quad 6<6.90<7

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 4.56 into 456.

4.56 \times 100 = 456

Times 100 moves the point two spots, a place-value shift.

Answer: 456

4 · Reviewdoes it hold up?

The digits 2, 3, 4, 5, 6 are consecutive and single-digit. The sum 6.90 is indeed between 6 and 7. Multiplying 4.56 by 100 gives 456, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (6, 7); only A=2 survives, giving 100 x 4.56 = 456.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 2.34 + 4.56 and multiplying 4.56 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 2.34 and 4.56 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 4 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $9$ and less than $10$ , find $100$ times $A.BC$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 9 and 10. Find 100 times A.BC.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
9 < A.BC + C.DE < 10.

Unknowns

The value of 100 x A.BC.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 9 and 10.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 9 and 10

#6 Guess and Check 5.NBT.A.3

We need the sum near 9, so 2A+2 should be about 9. Test A=3: the digits are 3, 4, 5, 6, 7, so the decimals are A.BC = 3.45 and C.DE = 5.67.

2A+2 \approx 9 \Rightarrow A=3 \Rightarrow A.BC=3.45,\ C.DE=5.67

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 3.45 + 5.67 = 9.12, which is greater than 9 and less than 10. So A=3 fits and A.BC = 3.45.

3.45+5.67=9.12,\quad 9<9.12<10

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 3.45 into 345.

3.45 \times 100 = 345

Times 100 moves the point two spots, a place-value shift.

Answer: 345

4 · Reviewdoes it hold up?

The digits 3, 4, 5, 6, 7 are consecutive and single-digit. The sum 9.12 is indeed between 9 and 10. Multiplying 3.45 by 100 gives 345, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (9, 10); only A=3 survives, giving 100 x 3.45 = 345.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 3.45 + 5.67 and multiplying 3.45 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 3.45 and 5.67 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 5 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $11$ and less than $12$ , find $100$ times $A.BC$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 11 and 12. Find 100 times A.BC.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
11 < A.BC + C.DE < 12.

Unknowns

The value of 100 x A.BC.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 11 and 12.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 11 and 12

#6 Guess and Check 5.NBT.A.3

We need the sum near 11, so 2A+2 should be about 11. Test A=4: the digits are 4, 5, 6, 7, 8, so the decimals are A.BC = 4.56 and C.DE = 6.78.

2A+2 \approx 11 \Rightarrow A=4 \Rightarrow A.BC=4.56,\ C.DE=6.78

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 4.56 + 6.78 = 11.34, which is greater than 11 and less than 12. So A=4 fits and A.BC = 4.56.

4.56+6.78=11.34,\quad 11<11.34<12

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 4.56 into 456.

4.56 \times 100 = 456

Times 100 moves the point two spots, a place-value shift.

Answer: 456

4 · Reviewdoes it hold up?

The digits 4, 5, 6, 7, 8 are consecutive and single-digit. The sum 11.34 is indeed between 11 and 12. Multiplying 4.56 by 100 gives 456, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (11, 12); only A=4 survives, giving 100 x 4.56 = 456.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 4.56 + 6.78 and multiplying 4.56 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 4.56 and 6.78 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 6 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $4$ and less than $5$ , find $100$ times $C.DE$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 4 and 5. Find 100 times C.DE.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
4 < A.BC + C.DE < 5.

Unknowns

The value of 100 x C.DE.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 4 and 5.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 4 and 5

#6 Guess and Check 5.NBT.A.3

We need the sum near 4, so 2A+2 should be about 4. Test A=1: the digits are 1, 2, 3, 4, 5, so the decimals are A.BC = 1.23 and C.DE = 3.45.

2A+2 \approx 4 \Rightarrow A=1 \Rightarrow A.BC=1.23,\ C.DE=3.45

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 1.23 + 3.45 = 4.68, which is greater than 4 and less than 5. So A=1 fits and C.DE = 3.45.

1.23+3.45=4.68,\quad 4<4.68<5

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 3.45 into 345.

3.45 \times 100 = 345

Times 100 moves the point two spots, a place-value shift.

Answer: 345

4 · Reviewdoes it hold up?

The digits 1, 2, 3, 4, 5 are consecutive and single-digit. The sum 4.68 is indeed between 4 and 5. Multiplying 3.45 by 100 gives 345, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (4, 5); only A=1 survives, giving 100 x 3.45 = 345.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 1.23 + 3.45 and multiplying 3.45 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 1.23 and 3.45 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 7 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $11$ and less than $12$ , find $100$ times $C.DE$ .

Your answer

Show solution

1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 11 and 12. Find 100 times C.DE.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
11 < A.BC + C.DE < 12.

Unknowns

The value of 100 x C.DE.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 11 and 12.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 11 and 12

#6 Guess and Check 5.NBT.A.3

We need the sum near 11, so 2A+2 should be about 11. Test A=4: the digits are 4, 5, 6, 7, 8, so the decimals are A.BC = 4.56 and C.DE = 6.78.

2A+2 \approx 11 \Rightarrow A=4 \Rightarrow A.BC=4.56,\ C.DE=6.78

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 4.56 + 6.78 = 11.34, which is greater than 11 and less than 12. So A=4 fits and C.DE = 6.78.

4.56+6.78=11.34,\quad 11<11.34<12

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 6.78 into 678.

6.78 \times 100 = 678

Times 100 moves the point two spots, a place-value shift.

Answer: 678

4 · Reviewdoes it hold up?

The digits 4, 5, 6, 7, 8 are consecutive and single-digit. The sum 11.34 is indeed between 11 and 12. Multiplying 6.78 by 100 gives 678, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (11, 12); only A=4 survives, giving 100 x 6.78 = 678.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 4.56 + 6.78 and multiplying 6.78 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 4.56 and 6.78 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Variant 8 medium

Five consecutive one-digit whole numbers, written in order from smallest to largest, are $A$ , $B$ , $C$ , $D$ , $E$ . The decimals $A.BC$ and $C.DE$ are formed from these digits. If the sum $A.BC + C.DE$ is greater than $9$ and less than $10$ , find $100$ times $C.DE$ .

Your answer

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1 · Understandwhat's really being asked

Five one-digit whole numbers in a row (each one bigger than the last) are named A, B, C, D, E. Build the decimals A.BC and C.DE from those digits. Their sum is somewhere between 9 and 10. Find 100 times C.DE.

Givens

A, B, C, D, E are five consecutive whole numbers in increasing order, so B=A+1, C=A+2, D=A+3, E=A+4.
A.BC has whole part A and decimal digits B (tenths) and C (hundredths).
C.DE has whole part C and decimal digits D (tenths) and E (hundredths).
9 < A.BC + C.DE < 10.

Unknowns

The value of 100 x C.DE.

Constraints

All five digits are single digits (0-9) and consecutive.
The sum of the two decimals lies strictly between 9 and 10.

2 · Planchoose the strategy

#6 Guess and Check · also uses: #9 Solve an Easier Related Problem

Because the digits are consecutive, the whole choice is fixed by a single starting digit A. The whole-number part of the sum is A + C = A + (A+2) = 2A+2, which already tells us roughly how big the sum is, so we test the few possible A values.

3 · Execute4 carry out the plan

1See that everything depends on A

#9 Solve an Easier Related Problem 5.NBT.A.1

Since the numbers are consecutive, once A is chosen the digits are A, A+1, A+2, A+3, A+4. The whole parts of A.BC and C.DE are A and C = A+2, so the whole part of the sum is about A + (A+2) = 2A+2.

A+C = A+(A+2) = 2A+2

Reducing five unknown digits to one starting number makes the problem easy to test.

2Find which A lands the sum between 9 and 10

#6 Guess and Check 5.NBT.A.3

We need the sum near 9, so 2A+2 should be about 9. Test A=3: the digits are 3, 4, 5, 6, 7, so the decimals are A.BC = 3.45 and C.DE = 5.67.

2A+2 \approx 9 \Rightarrow A=3 \Rightarrow A.BC=3.45,\ C.DE=5.67

A guess pinned by the whole-number size is quick to verify.

3Check the sum condition

#6 Guess and Check 5.NBT.B.7

Add the two decimals: 3.45 + 5.67 = 9.12, which is greater than 9 and less than 10. So A=3 fits and C.DE = 5.67.

3.45+5.67=9.12,\quad 9<9.12<10

Only one starting digit makes the sum sit in the window.

4Multiply by 100

#6 Guess and Check 5.NBT.B.7

Multiplying a hundredths decimal by 100 shifts the decimal point two places right, turning 5.67 into 567.

5.67 \times 100 = 567

Times 100 moves the point two spots, a place-value shift.

Answer: 567

4 · Reviewdoes it hold up?

The digits 3, 4, 5, 6, 7 are consecutive and single-digit. The sum 9.12 is indeed between 9 and 10. Multiplying 5.67 by 100 gives 567, a whole number, which makes sense because two decimal places cleared exactly.

Another way: Systematic List: tabulate the sum for each possible A (0 -> 2.46; 1 -> 4.68; 2 -> 6.90; 3 -> 9.12; 4 -> 11.34; 5 -> 13.56) and keep only the row inside (9, 10); only A=3 survives, giving 100 x 5.67 = 567.

Standardsmin grade 5

5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths — Adding 3.45 + 5.67 and multiplying 5.67 x 100.
5.NBT.A.1 Recognize that a digit in one place represents ten times as much as to its right — Reducing the five consecutive digits to a single starting digit A.
5.NBT.A.3 Read, write, and compare decimals to thousandths — Forming the decimals 3.45 and 5.67 from the digit list.

💡Takeaway. When the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!

Place consecutive whole numbers to sum decimals · 8 practice problems

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