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if a two digit positive integer has its digits reversed, the sum of

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if a two digit positive integer has its digits reversed, the sum of  [#permalink]

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New post 28 Dec 2017, 08:51
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if a two digit positive integer has its digits reversed, the sum of the resulting integer and its original integer is 110. What is the sum of the digits in the original number?

A) 7
B) 8
C) 10
D) 11
E) 12

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Re: if a two digit positive integer has its digits reversed, the sum of  [#permalink]

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New post 28 Dec 2017, 08:59
saswata4s wrote:
if a two digit positive integer has its digits reversed, the sum of the resulting integer and its original integer is 110. What is the sum of the digits in the original number?

A) 7
B) 8
C) 10
D) 11
E) 12


Any two digit integer can be expressed as 10a + b, where a and b are tens and units digits, respectively. For example, 23 = 2*10 + 3.

Given that (10a + b) + (10b + a) = 110 --> 11a + 11b = 110 --> a + b = 10.

Answer: C.
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Re: if a two digit positive integer has its digits reversed, the sum of  [#permalink]

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New post 28 Dec 2017, 09:00
saswata4s wrote:
if a two digit positive integer has its digits reversed, the sum of the resulting integer and its original integer is 110. What is the sum of the digits in the original number?

A) 7
B) 8
C) 10
D) 11
E) 12


let number be ab..
so number = 10a+b and reverse = 10b +a..
sum = \(11a+11b=110...a+b=\frac{110}{11}=10\)

C
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if a two digit positive integer has its digits reversed, the sum of  [#permalink]

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New post 28 Dec 2017, 19:41
Let the original number be represented as xy, thus its value =10*x + y
The new number after the digits have been reversed would be yx, and the value = 10*y + x

Sum of values of original & new number = 10*x+y+10*y+x
= 11*x+11*y
=11*(x+y)

But we know the sum is 110
So. 11*(x+y) = 110
x+y =10 => sum of digits (x,y)
if a two digit positive integer has its digits reversed, the sum of &nbs [#permalink] 28 Dec 2017, 19:41
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