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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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05 Sep 2013, 21:24
Bunuel wrote: u2lover wrote: Each digit in the twodigit number G is halved to form a new twodigit number H. Which of the following could be the sum of G and H?
A. 153 B. 150 C. 137 D. 129 E. 89 Twostep solution: G + G/2 = 3G/2 > the sum is a multiple of 3. G is a twodigit 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 ?



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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05 Sep 2013, 23:51
ygdrasil24 wrote: Bunuel wrote: u2lover wrote: Each digit in the twodigit number G is halved to form a new twodigit number H. Which of the following could be the sum of G and H?
A. 153 B. 150 C. 137 D. 129 E. 89 Twostep solution: G + G/2 = 3G/2 > the sum is a multiple of 3. G is a twodigit 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 twodigit number. Hope it's clear.
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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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06 Sep 2013, 00:08
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 twodigit number. Hope it's clear.[/quote] Yes it is thanks So basically G ranges from 20 to 198 for all G >0



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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06 Sep 2013, 00:12



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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06 Sep 2013, 00:21
Bunuel wrote: ygdrasil24 wrote: Yes it is thanks So basically G ranges from 20 to 198 for all G >0 No. G must also be a two digit number, so it ranges from 20 to 88. Hmm... blunder as always By the way why cant G(max) be 98 , H(max) be 49 in that case



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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06 Sep 2013, 00:23



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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06 Sep 2013, 00:26
By the way why cant G(max) be 98 , H(max) be 49 in that case[/quote]
We are told that EACH digit in the twodigit number G is halved, thus both digits of G must be even.[/quote]
Hmmm..Okay under even constraint G max =88, Thanks



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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10 Sep 2013, 22:03
I followed the approach as :
Multiplied each number with 2/3 and saw only 129 gives a 2 digit number i.e 43+86 which is possible,
for all of the rest number it gives a 3 digit number or is not multiple of 3.
Thanks



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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21 Nov 2013, 06:08
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.



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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04 Mar 2014, 17:14
Let H be the 2digit 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 singledigit 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 twodigit number G is halved to form a new [#permalink]
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04 Mar 2014, 19:45
Maximum largest number = 88 + 44 = 132, so options A,B & C are eliminated Number should be divisible by 3 (for ex a + a/2 = 3a/2) 129 > Divisible by 3 >>>>>>>>>>>> Answer = D89 > Not divisible by 3
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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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04 Mar 2014, 19:48
prsnt11 wrote: Let H be the 2digit 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 singledigit 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. C is also eliminated
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Each digit in the twodigit number G is halved to form a new [#permalink]
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13 Mar 2015, 03:11
u2lover wrote: Each digit in the twodigit number G is halved to form a new twodigit number H. Which of the following could be the sum of G and H?
A. 153 B. 150 C. 137 D. 129 E. 89 let G = 10a+b => H = (10a+b)/2 G+H = (3/2)*(10a+b) out of the options C and E go out as they are not divisible by 3. substitute other options and try... 1) 153 = 3/2*G => G = a three digit number (goes out) 2) G = a three digit number (goes out) 4) no need to check but just to be sure... G = 86 H = 43 G+H = 129 correct. takes less than a minute to solve....
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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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28 May 2015, 21:24
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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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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23 Dec 2015, 19:08
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.



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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28 Dec 2015, 12:56
G = 10x + y H = 10 (x/2) + (y/2) Adding the two, the number is in the form 1.5 (10x + y) Look for options that can be represented in this format and is also a two digit number. Only D satisfies the condition. Hence, D
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Each digit in the twodigit number G is halved to form a new [#permalink]
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12 Mar 2016, 16:21
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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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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12 Jan 2017, 13:48
Let G = 10a + b hence [halved G] = H = 10a/2 + b/2 = 5a + 3/2
Then G + H = 15a + 3/2b . Multiplying both sides by 2 we get 2(G+H) = 30a + 3b = 3(10a+b)
This means that (G+H) HAS TO BE A MULTIPLE OF 3 and this is fulfilled by answer D.



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Re: Each digit in the twodigit number G is halved to form a new [#permalink]
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22 Feb 2017, 23:55
There is a way of saving 60% time by eliminating the first three answers (A, B & C): Since both digits of G must be even (so as to yield an integer when halved), the highest possible value of G is 88. So the highest possible value of G+H is 88+44=132 which eliminates the first three answers. So, we now have to test only the last two answers to find out which satisfies the equation 3G/2 for an integer value of G.




Re: Each digit in the twodigit number G is halved to form a new
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