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# If s is a two-digit number (so s = qp with q and p digits),

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If s is a two-digit number (so s = qp with q and p digits), [#permalink]

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31 Jan 2005, 06:21
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If s is a two-digit number (so s = qp with q and p digits), what is the last digit p of s?

(1) The number 3s is a three-digit number whose last digit is p
(2) The digit p is less than 7

(A) statement 1 alone is sufficient to answer the question, but statement 2 alone is not sufficient
(B) statement 2 alone is sufficient to answer the question, but statement 1 alone is not sufficient
(C) both statements together are needed to answer the question, but neither statement alone is sufficient
(D) either statement by itself is sufficient to answer the question
(E) not enough facts are given to answer the question
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31 Jan 2005, 08:09
If s is a two-digit number (so s = qp with q and p digits), what is the last digit p of s?

(1) The number 3s is a three-digit number whose last digit is p
=> q>3 & p is either 0 or 1 or 5.

(2) The digit p is less than 7
=> p can be 1, 2.....6

(1)&(2) => p could be either 0 or 1 or 5.

Hence, E.
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31 Jan 2005, 08:21
Agree, (E). p could be 0 or 5 for (I).
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31 Jan 2005, 08:28
speedo wrote:
If s is a two-digit number (so s = qp with q and p digits), what is the last digit p of s?

(1) The number 3s is a three-digit number whose last digit is p
=> q>3 & p is either 0 or 1 or 5.

(2) The digit p is less than 7
=> p can be 1, 2.....6

(1)&(2) => p could be either 0 or 1 or 5.

Hence, E.

"E" for me.

Speedo, p can't be 1, it can only be 0 or 5.
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31 Jan 2005, 20:30
E.

But, Can some one explain why for St.1, p should be 0 or 5?.
Is it 3*S(a 2 digit number) or the digit 3 precedes S?.
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03 Feb 2005, 05:58
clearly E
At first I've made the same mistake than Speedo so it's better to take time because you always make that kind of stupid error when you read too fast...

prep_gmat, the statement 1 just gives us 2 potential numbers : 0 and 5
ex : 10*3 = 30 (last digit of 3S = P = 0)
ex : 15*3 = 45 (last digit of 3S = P = 5)

you can find it fast :

0*3=0
1*3=3
2*3=6
3*3=9
4*3=12
5*3=15
6*3=18
7*3=21
8*3=24
9*3=27

just 2 numbers have the same last digit after being multiplied by 3, number 0 and number 5.

Last edited by Antmavel on 03 Feb 2005, 17:44, edited 1 time in total.
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03 Feb 2005, 10:41
Yes, it must be 0 or 5 for St.1. My mind had gone a walk about...
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