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nick1816
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Find the sum of digits of smallest positive integer that is both 28% l 12 Apr 2020, 02:23
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D
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95% (hard)

Question Stats:

40% (02:38) correct 60% (02:40) wrong based on 91 sessions

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Find the sum of digits of smallest positive integer that is both 28% larger than some positive integer and 28% smaller than another positive integer.

A. 5
B. 9
C. 13
D. 18
E. 20
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D
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Aviralft9
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Find the sum of digits of smallest positive integer that is both 28% l Updated on: 12 Apr 2020, 05:16
Originally posted by Aviralft9 on 12 Apr 2020, 04:37.
Last edited by Aviralft9 on 12 Apr 2020, 05:16, edited 1 time in total.
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Let K be the required number, and X be the number that is 28% more and Y be the number that is 28% less than K;
So, K = (128/100) * X = (72/100) * Y;

(128/100) * X = (72/100) * Y;
16/9 = Y/X;

Assuming Y = 16;
Find the smallest integer for P, such that 16*P(72/100) is a integer.
P =25;
therefore the X = 225 and Y = 400;
=> 72% of 400 = 288 is the required integer.
Sum of digits = 18;
IMO D
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Find the sum of digits of smallest positive integer that is both 28% l 12 Apr 2020, 05:08
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Aviralft9
Let K be the required number, and X be the number that is 28% more and Y be the number that is 28% less than K;
So, K = (128/100) * X = (72/100) * Y;

(128/100) * X = (72/100) * Y;
16/9 = Y/X;

Assuming Y = 16;
Find the smallest integer for P, such that 6*P(72/100) is a integer.
P =100;
therefore the X = 900 and Y = 1600;
=> 72% of 1600 = 1152 is the required integer.
Sum of digits = 9;
IMO B
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I may be wrong brother but I got the smallest positive integer as 288 and sum of the digits is 18 so IMO answer is d

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Aviralft9
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Find the sum of digits of smallest positive integer that is both 28% l 12 Apr 2020, 05:14
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Peddi
Aviralft9
Let K be the required number, and X be the number that is 28% more and Y be the number that is 28% less than K;
So, K = (128/100) * X = (72/100) * Y;

(128/100) * X = (72/100) * Y;
16/9 = Y/X;

Assuming Y = 16;
Find the smallest integer for P, such that 6*P(72/100) is a integer.
P =100;
therefore the X = 900 and Y = 1600;
=> 72% of 1600 = 1152 is the required integer.
Sum of digits = 9;
IMO B
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I may be wrong brother but I got the smallest positive integer as 288 and sum of the digits is 18 so IMO answer is d

Hello Peddi,
Could you explain how did you get it? After having a second look at it, I realized D is the correct ans.
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Find the sum of digits of smallest positive integer that is both 28% l 12 Apr 2020, 05:49
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Aviralft9
Peddi
Aviralft9
Let K be the required number, and X be the number that is 28% more and Y be the number that is 28% less than K;
So, K = (128/100) * X = (72/100) * Y;

(128/100) * X = (72/100) * Y;
16/9 = Y/X;

Assuming Y = 16;
Find the smallest integer for P, such that 6*P(72/100) is a integer.
P =100;
therefore the X = 900 and Y = 1600;
=> 72% of 1600 = 1152 is the required integer.
Sum of digits = 9;
IMO B
I may be wrong brother but I got the smallest positive integer as 288 and sum of the digits is 18 so IMO answer is d

Hello Peddi,
Could you explain how did you get it? After having a second look at it, I realized D is the correct ans.
Show more
You have done everything right until y/x= 16/9 , since in question it is mentioned 28 percent 28/100 translates to 7/25 , so the integers which are supposed to be 28 percent more or 28 percent less are supposed to be multiples of 25 , so since the lowest multiple of 25 is 25 itself the min value of y if 25*16 and min value of x is 25*9 , so the integer which is which is 28percent more than 225 and 28p less than 400 is 288 so the sum of the digits =2+8+8=18

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Find the sum of digits of smallest positive integer that is both 28% l 23 Dec 2025, 22:28
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Dis ones a bit difficult to figure out wut its saying.......so....lets get into it........

Let.....a number is.... X .....and 28 % larger than X is...... Y ......so.... Y = X [ 128 / 100 ] = X [ 32 / 25 ]
and then....this number Y .....will be 28 % smaller than another number Z.....so..... Y = Z [ 72 / 100 ] = Z [ 18 / 25 ]
So..... Y = Z [ 18 / 25 ] ....in other words..... Y [ 25 / 18 ] = Z .......

So.....X [ 32 / 25 ] = Y ......and....... Y [ 25 / 18 ] = Z ......
So...we gotta take a minimum number.... 32X ..... that is divisible by 25 and 18 both......

So.... 32X = [ 2^5 ] X ..... but 25 = 5^2 .....and 5 is prime and 5 isn't a factor of 32..... so X must have at least 25 as a factor.....then 18 = 2 × 3 × 3 ..... and there is a 2 inside 32 .... but 3 ain't a factor of 32..... so X must also have at least........
[ 3 × 3 ] = 9 as a factor.......
So....X must have at least 25 and 9 as factor.......so X must be at least = 25 × 9 = 225 ......

So...if X = 225 ..... then Y = 225 × [ 32 / 25 ] = 288 ......

So.... Y = 288...... is da number dat is 28 % bigger than X ....and..... 28 % smaller then Z ......

So...sum of digits of Y is ..... 2 + 8 + 8 = 18 .......

! nah id win!
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