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Each digit in the two-digit number G is halved to form a new two-digit number H. Which of the following could be the sum of G and H?

A. 153 B. 150 C. 137 D. 129 E. 89

Two-step solution:

G + G/2 = 3G/2 --> the sum is a multiple of 3.

G is a two-digit number --> G < 100 --> 3G/2 < 150.

Among the answer choices the only multiple of 3 which is less than 150 is 129.

Answer: D.

What could be the minimum number ?

Assuming G is a positive number, the least value of G+G/2 will be 20+10=30. G must be even and cannot be less that 20. If it's an even number less than 20, then G/2 will not be a two-digit number.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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05 Sep 2013, 23:08

What could be the minimum number ?[/quote]

Assuming G is a positive number, the least value of G+G/2 will be 20+10=30. G must be even and cannot be less that 20. If it's an even number less than 20, then G/2 will not be a two-digit number.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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21 Nov 2013, 05:08

Let G be XX. Let H be x/2 x/2. G+H= 3 () Its a multiple of 3. Only two numbers fit the bill 153 and 129. 153/3 = 51 ( not possible because G is a two digit number and 51 is half of 102). Hence (D) 129.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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04 Mar 2014, 16:14

Let H be the 2-digit number xy (actually 10x+y). Then G must be 2x2y (actually 10(2x) + 2y or 2(10x+y). In other words, the digits of G must be even single-digit numbers. The maximum value of G can be 88 and thereby H can be 44. Therefore, maximum value of G+H = 132. Therefore, A & B are out. Now G+H = 3(10x+y) implies, G+H must be a multiple of 3. Only D among the remaining answer choices is a multiple of 3. So D is the answer.

Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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04 Mar 2014, 18:48

prsnt11 wrote:

Let H be the 2-digit number xy (actually 10x+y). Then G must be 2x2y (actually 10(2x) + 2y or 2(10x+y). In other words, the digits of G must be even single-digit numbers. The maximum value of G can be 88 and thereby H can be 44. Therefore, maximum value of G+H = 132. Therefore, A & B are out. Now G+H = 3(10x+y) implies, G+H must be a multiple of 3. Only D among the remaining answer choices is a multiple of 3. So D is the answer.

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Re: Each digit in the two-digit number G is halved to form a new [#permalink]

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28 May 2015, 20:24

Each of digits in a 2 digit number when halved gives another number. This means both the digits must be even. Possible combinations are as follows: Digit Half sum ---------- ---- ----- 8 4 12 6 3 9 4 2 6 2 1 3

Looking at choices only 129 is the possible answer : numbers being 86 & 43.

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23 Dec 2015, 18:08

so, the digits of G must be even. suppose G = 10A+B where A- tens digit, and B units digit. now we are told that 5A+0.5B=H.

we are asked for the sum of G and H, or 15A+1.5B

knowing that A and B must be even, there aren't that many ways it can be arranged in that way. 4*4 = 16 possibilities. nevertheless, since the results are over 100, we can definitely rule out 2, 4, and 6 as A. Thus, A must be 8.

Now, we have 15*8 = 120. We're getting closer to the answer choice. B can be 2 = so the min digit of H can be 1. B can be 8 = so the max digit of H can be 4.

suppose B = 8 => 1.5B = 12, and the sum of G+H=132. suppose B = 2 => 1.5B = 3, and the sum of G+H=123.

now we know for sure that G+H must be a number between 123 and 132. only one answer choice satisfies this condition.

Each digit in the two-digit number G is halved to form a new [#permalink]

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12 Mar 2016, 15:21

G=10a+b H=10c+d, here c=a/2 & d=b/2 or a=2c and b=2d substitute these values G=20c+2d; G+H =30c+3d = 3(10c+d) now in options check for which options are divisible by 3; options A, B and D are divisible by 3; A. 153 => 153/3=51 => c cannot be 5 as d will be a single digit integer. B. 150 => 150/3=50 => c cannot be 5 as d will be a single digit integer. D. 129 => 129/3=43 ---- CORRECT ANS. option D

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