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number theory

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by sharmishtha_goel » Wed Aug 10, 2011 6:55 am
For any four digit number, abcd, *abcd*= (3a)(5b)(7c)(11d). What is the value of (n - m) if m and n are four-digit numbers for which *m* = (3r)(5s)(7t)(11u) and *n* = (25)(*m*)?

2000
200
25
20
2


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by kmittal82 » Wed Aug 10, 2011 7:12 am
Something's not right with the question, can you please double check?
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by Frankenstein » Wed Aug 10, 2011 7:16 am
kmittal82 wrote:Something's not right with the question, can you please double check?
Hi,
Yes, I do feel the same. I think it should be *abcd*= (3^a)(5^b)(7^c)(11^d) and *m* = (3^r)(5^s)(7^t)(11^u).
So, if m = rstu
n should be r(s+2)tu.
So, the difference will be 200.
Cheers!

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by sharmishtha_goel » Wed Aug 10, 2011 8:30 am
Frankenstein wrote:
kmittal82 wrote:Something's not right with the question, can you please double check?
Hi,
Yes, I do feel the same. I think it should be *abcd*= (3^a)(5^b)(7^c)(11^d) and *m* = (3^r)(5^s)(7^t)(11^u).
So, if m = rstu
n should be r(s+2)tu.
So, the difference will be 200.

Oh !! My bad!!!
It is exactly what you have written it to be!!

My doubt is, is the no. m was : (3^1)(5^1)(7^1)(11^1)
n = (3^1)(5^3)(7^1)(11^1)
n-m = 28875 -1155
!= 200 :(

what am i assuming incorrect :(

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by Frankenstein » Wed Aug 10, 2011 8:36 am
sharmishtha_goel wrote: Oh !! My bad!!!
It is exactly what you have written it to be!!

My doubt is, is the no. m was : (3^1)(5^1)(7^1)(11^1)
n = (3^1)(5^3)(7^1)(11^1)
n-m = 28875 -1155
!= 200 :(

what am i assuming incorrect :(

TIA
Hi,
The problem with your approach is that it is given *m* = (3^1)(5^1)(7^1)(11^1)
So, m = 1111
Now, *n* = (25)*m* = (3^1)(5^1)(7^1)(11^1)(5^2) = (3^1)(5^3)(7^1)(11^1).
So, n = 1311
So, n-m = 1311 - 1111 = 200
We need n-m , not *n* - *m*.
For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d)
Cheers!

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