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Re: odds and evens!
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05 May 2011, 08:35
AnkitK wrote: Each digit in the 2 digit number G is halved to form a new 2 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 Say H=xy [x is the ten's place digit after being halved and y is the unit's place digit after being halved] H=10x+y G=(2x)(2y) G=10*(2x)+2y Possible values of 2x=2,4,6,8; 2y=0,2,4,6,8 Possible values of x=1,2,3,4; y=0,1,2,3,4 G+H=20x+2y+10x+y=30x+3y=3(10x+y) Thus, the sum must be a multiple of 3; options C and E are out. Let's try other options: A. 3(10x+y)=153 i.e. 10x+y=51 i.e. 10*5+1=51; x=5(Not a possible value of x) and y=1 B. 3(10x+y)=150 i.e. 10x+y=50 i.e. 10*5+0=50; x=5(Not a possible value of x) and y=0 D. 3(10x+y)=129 i.e. 10x+y=43 i.e. 10*4+3=43; x=4(Possible) and y=3(Possible) Ans: D
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Re: odds and evens!
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05 May 2011, 16:23
G = 10x + y H = 5x + y/2 so G + H = 15x + 3/2y. Multiply by 2 and you get 2 ( G+H )= 3 (10 x + y) so G+H must be a multiple of 3 and G a multiple of 2 (obviously otherwise we cant divide G) and we know G less than or equal to 88 (highest 2 digits even) and so H less than or equal to 44 (half G), so G+H less than 132. ABC out and E out because not a multiple of 3 Answer is D Hope this is helpful
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Re: odds and evens!
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06 May 2011, 15:49
G is even
H can be even or odd
G's unit digit possibilities are 0 2 4 6 8
H's unit digit possibilities are 0 1 2 3 4
No checking the answers we can see only D fits the criteria.
Answer is D.



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Re: Each digit in the twodigit number G is halved to form a new
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13 Aug 2013, 10:18
let the number be x, now x is halved => x/2..the question asks what is x? x(number) + x/2 (half of that number) = something => 3x/2 = something => x = 2 (something) / 3 so the answer choice must be divisible by 3 eliminate C, E right away check the rest D satisfies 3x/ 2 = 129 => x = 86 Verify. 86 + 43 = 129 Answer is D
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Re: Each digit in the twodigit number G is halved to form a new
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13 Aug 2013, 13:18
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
please explain My thinking about this problem (detailed explanation): use this flow chart
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Re: Each digit in the twodigit number G is halved to form a new
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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
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05 Sep 2013, 23:51



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




Re: Each digit in the twodigit number G is halved to form a new &nbs
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28 May 2015, 21:24



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