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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


when W is divided by 14, the reminder is 0. if W is three lesser than it value and when divided by 15 its remainder is 14. what is the value of W ?

a. 182
b.282
c. 382
d.482
e. 582




Since W is divisible by 14 we can put W=14k.
Similarly we can put W-3=15t+14 ---> W=15(t+1)+2.

So we have 14k=15(t+1)+2. Let t+1 be T.
Then we have 14k-2 = 15T. Since 14k-2 is divisible by 2 but 15 isn't, T should be divisible by 2. So let T be 2l.
Then we have 7k-15l=1. We know that 7*(-2) + 15*1 =1.
So we may put 1=7*(15A-2) +15*(-7A+1).

Let A=1 then we have k=15*(1)-2=13. So W= 14*13=182.

The answer is, therefore, (A).
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anik1989
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here only 182 is divisible by 14
so it should be A


solving using the data in the question-

as W is divisible by 14
w =14x (where x can be some positive integer)
also
w-3=15P+14 => W=15p+17
subtract 17 from the answer choices 182-17=165 which is divisible by 15 so A

thank you very much. but one more step ahead, i want to know , if i proceed this problem without any option, then what should be my approach?


Hi,
incase you do not have choices or you do not want to substitute values, i would think of following way as a quick method of finding the answer..
W is div by 14..
W-3, when divided by 15 gives you 14 as remainder..
this means if i add 1 to W-3, the remainder will increase to 15 or W-2 is div by 15..
in other words W would leave a remainder of 2, when div by 15..

armed with the info that W leaves a remainder of 0, when div by 14 and 2, when div by 15
, we can move to next step...
first we find the number that is div by both and that would be the LCM of 14 and 15..
so 14 * 15 would leave a remainder 0, when div by 14 or 15..
when we take a lower multiple of 14, it would leave a remainder equal to the diff of 15 and 14, which is 1..
so for the remainder to be 2, when divided by 15... the 2nd lower multiple of 14 is required, which is 14*(15-2)=14*13
and 14*13=182.. the answer A
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


when W is divided by 14, the reminder is 0. if W is three lesser than it value and when divided by 15 its remainder is 14. what is the value of W ?

a. 182
b.282
c. 382
d.482
e. 582




Since W is divisible by 14 we can put W=14k.
Similarly we can put W-3=15t+14 ---> W=15(t+1)+2.

So we have 14k=15(t+1)+2. Let t+1 be T.
Then we have 14k-2 = 15T. Since 14k-2 is divisible by 2 but 15 isn't, T should be divisible by 2. So let T be 2l.
Then we have 7k-15l=1. We know that 7*(-2) + 15*1 =1.
So we may put 1=7*(15A-2) +15*(-7A+1).

Let A=1 then we have k=15*(1)-2=13. So W= 14*13=182.

The answer is, therefore, (A).


Not able to understand the steps in bold letter , please explain.
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anik1989
when W is divided by 14, the reminder is 0. if W is three lesser than it value and when divided by 15 its remainder is 14. what is the value of W ?

a. 182
b.282
c. 382
d.482
e. 582

is there any method to calculate directly rather than putting each choice ??

The basics of the method are discussed in this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... unraveled/

W when divided by 14 is completely divisible. This means you can make groups of 14 out of W.

Let's put one group of 14 aside (since it will act as the remainder) and consider the other groups of 14.
Now if we take away 3 from another group of 14 (11 will be leftover), the total will be divisible by 15.

(14a + 11) is divisible by 15.

If a = 11, (14*11 + 11) = 15 * 11 = 165 will be divisible by 15.

To get W, we add back the 3 we removed and the group of 14 that we excluded initially to get 165 + 3 + 14 = 182 = W

Answer (A)
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anik1989
when W is divided by 14, the reminder is 0. if W is three lesser than it value and when divided by 15 its remainder is 14. what is the value of W ?

a. 182
b.282
c. 382
d.482
e. 582

is there any method to calculate directly rather than putting each choice ??

Hello Experts,

If w = 14k which means w is a multiple of 14 and it has to contain 2 and 7 atleast or it has to be divisible by 2 and 7 both. If we scan the answer choices we can see that all five options are divisible by 2 since the units digit of all 5 answer choices are even and only 182 is divisible by 7 and rest all 4 answer choices are not divisible by 7 so from this much solving we can conclude that 182 is the right answer. Is this method correct ?
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