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Number Properties ( Intermediate Level )

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by Aman verma » Tue Aug 09, 2011 9:23 am
Q: If X = abc and Y = uv are 3 digits and 2 digits natural numbers respectively, such that u and v must be distinct integers, then how many pairs of X and Y are there in total which gives the same result when we multiply abc with uv as the product of cba with vu ( i.e the position of digits is interchanged).


a) 6

b) 7

c) 8

d) 9

e) 10
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by GmatKiss » Tue Aug 09, 2011 9:28 am
Tough one! what is the source?
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by gmatboost » Tue Aug 09, 2011 9:57 pm
This question is both way too hard/complex (in my opinion), and also flawed as stated, since there are at least 12 solutions.

253*64
352*46
(this is two different pairs of X, Y, though they are the reverse of each other)

693*24
396*42

286*62
682*26

143*62
341*26

132*63
231*36

132*84
231*48

It's not worth going into where these came from, but let's just say it took a while. And I am pretty sure there are more out there.
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by Aman verma » Thu Aug 11, 2011 4:45 am
Thanks Gmatboost for the answer. Now the pair 253*64 & 352*46 constitutes a single pair.

Similarly:

693*24
396*42

286*62
682*26

143*62
341*26

132*63
231*36

132*84
231*48

132*42
231*24

These all constitute single pair.I have added 1 pair to the list given by Gmatboost. There are 7 pairs in all that is given above and one pair is still missing. I am still working on it and I am pretty sure there is somebody out there who can find the missing pair. The OA is[spoiler]c) 8[/spoiler]
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by Frankenstein » Thu Aug 11, 2011 5:11 am
Hi,
In fact there are more cases such as:

154*82
451*28

143*93
341*39

143*31
341*13

264*42
462*24
.
.
.

Concept is same for all numbers: abc and uv
c/a = u/v and b = a+c

I guess we can ignore such questions.
Cheers!

Things are not what they appear to be... nor are they otherwise
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