# MATHEMATICS

The digit in the ten's place of a two digit number is one less than the digit in the one's place. If we add this number to the number obtained by reversing its digits, the result is 55, find the number. Pls answer quickly.

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1. 32 + 23 = 55

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2. Your wo digit number is:

10 a + b

The number obtained by reversing its digits is:

10 b + a

If we add this number to the number obtained by reversing its digits, the result is 55 means:

10 a + b + 10 b + a = 55

11 a + 11 b = 55

11 ( a + b ) = 55

Divide both sides by 11

a + b = 5

The digit in the ten's place of a two digit number is one less than the digit in the one's place so b can be:

b = 0

b = 1

or

b = 2

b can not be 3 , 4 or 5 , because in this case the digit in the ten's place is greater than the digit in the one's place.

Possible combinations are:

a = 5 , b = 0

a = 4 , b = 1

a = 3 , b = 2

For:

a = 5 , b = 0

Original number is:

10 a + b = 10 ∙ 5 + 0 = 50 + 0 = 50

The number obtained by reversing its digits is:

05 = 5

50 + 5 = 55

For:

a = 4 , b = 1

Original number is:

10 a + b = 10 ∙ 4 + 1 = 40 + 1 = 41

The number obtained by reversing its digits is:

14

41 + 14 = 55

For:

a = 3 , b = 2

Original number is:

10 a + b = 10 ∙ 3 + 2 = 30 + 2 = 32

The number obtained by reversing its digits is:

23

32 + 23 = 55

Possible solutions are:

50 , 41 , 32

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3. Only for 32, the digit in the ten's place of is one less than the digit in the one's place.

So final solution is:

32

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4. Do you mean 23 since the number is the 10s place is 1 less than the number in the ones place so 23 + 32 = 55 which agrees with PsyDag

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DrBob222

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