Competition Mode Question

If the difference between a two digit positive integer and its reversal is three times the sum of its digits, how many such pairs are there?

A. 1
B. 2
C. 3
D. 4
E. 5

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If the difference between a two digit positive integer and its reversal is three times the sum of its digits, how many such pairs are there?

A. 1
B. 2
C. 3
D. 4
E. 5

Let the two digits number be 10a + b
Reverse is 10b + a

Now, 10a + b - 10b - a = 3(a + b)
9a - 9b = 3a + 3b
6a = 12b
a = 2b

'a' can have only four values for which 'b' would have four values. The numbers are:
a = 1, b = 2
a = 2, b = 4
a = 3, b = 6
a = 4, b = 8

Hence only four numbers are possible.

IMO Answer D.
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10A+B-(10B+A)=3(A+B)
9(A-B)=3(A+B)
A=2B
0<A≤9
A={2,4,6,8}; B={1,2,4,3}
(A,B)={2,1;4,2;6,4;8,3}: 4 pairs

Ans (D)

Let the two digit number be m = xy = 10x + y
And number obtained by reversing be n = yx = 10y + x

Difference = 10x + y - (10y + x) = 9x - 9y = 9(x - y)

Given, 9(x - y) = 3(x + y)
—> 3(x - y) = x + y
—> 2x = 4y
—> x = 2y
—> Possible values = 12, 24, 36, 48
—> Possible pairs of 2 digit number and its reversal = {(12,21). (24,42), (36,63), (48,84)}
—> 4 pairs

IMO Option D

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Let the two digit positive number =ab
then its reverse is ba
ab-ba=3(a+b)
but ab=10a+b and ba=10b+a
10a+b-(10b+a)=3a+3b
9a-9b=3a+3b
6a-12b=0
6(a-2b)=0
This implies a=2b
since when b=0 a=0, b cannot be 0
only four other possible values exist which yield unique value for a, since the largest unit digit that a can take is 8 based on the given condition of a=2b.
The set of possible values of b={1, 2, 3, 4}

The answer is, therefore, option D in my view.

We can let a = tens digit and b = ones digit, and create the equation:

10a + b - (10b + a) = 3(a + b)

10a + b - 10b - a = 3a + 3b

6a = 12b

a = 2b

We see that a can be 2, 4, 6, or 8, so there are 4 such pairs.

Answer: D
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