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# Can Any one provide assistance??

Author Message
Intern
Joined: 18 Jan 2012
Posts: 1
Location: Pakistan
Concentration: Finance, Social Entrepreneurship
GMAT Date: 08-01-2014
GPA: 3.34
WE: Education (Education)
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Can Any one provide assistance?? [#permalink]

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18 Jan 2012, 08:10
I have some problems, if anyone knows the solution please guide me, thanks.

1. A number being successively divided by 3,5 and 8 leaves remainder 1,4 and 7 respectively. Find respective remainders if the orders of divisors is reversed?

2. When a certain number is multiplied by 13,the product consists entirely of fives. The smallest such number is:
a.41625 b.42135 c.42515 d.42735
(i do know the answer is d, but dn't know how)

3. The least number by which 72 must be multiplied in order to produce a multiple of 112, is:
a.6 b.12 c.14 d.18

4. A number when divided by 3 leaves a remainder 1. When the quotient is divided by 2, it leaves a remainder1. What will be the remainder when the number is divided by 6.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
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Re: Can Any one provide assistance?? [#permalink]

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31 Jan 2012, 14:44
I'm happy to contribute here.

This post is a real doozie! Four tough questions.

First, I will say: Questions #1 & #2, I believe, are right out as far as the GMAT -- I think they are much harder than anything the GMAT will expect you to do.

Question #3 is a perfectly legitimate GMAT question, a bit on the harder side, but well within expectations.

Question #4 could be on the GMAT -- it's at the harder end of what might be possible.
====================================================================================

Question #1:
A number being successively divided by 3,5 and 8 leaves remainder 1,4 and 7 respectively. Find respective remainders if the orders of divisors is reversed?

Well, I'm going to go in reverse order. Let's say q is the final quotient. On the final division, we divided by 8 and got a remainder of 7, so the number we had before that was: 8q + 7

That, in turn, was the quotient of the previous division, where we divided by 5 with a remainder of 4. Before we divided, we must have had: 5(8q + 7) + 4.

That, in turn, was the quotient of the previous division, where we divided by 3 with a remainder of 1. Before we divided, we must have had: 3[5(8q + 7) + 4] + 1 = N, the original dividend.

I purposely left that unmultiplied out, so you could see each step's contribution. Now, multiplying out, we have N = 240q + 118.

Now, divide in the reverse order --- obviously, 3, 5, and 8 all go into 240q, so the real question is what happens when we divide the aggregate remainder term, viz, 118.

118/8 = 14, remainder = 6

14/5 = 2, remainder = 4

2/3 = 0, remainder = 2

I believe {6, 4, 2} is the remainder chain when you divide with the chain of divisors in the opposite order.

Question #2:
2. When a certain number is multiplied by 13,the product consists entirely of fives. The smallest such number is:
a. 41625
b. 42135
c. 42515
d. 42735
e. none of the above

Well, start with the units digit --- of course, that has to be 5, so get a five in the unit's digit of the product. 5 x 13 = 65

Well, the next step is to get a 5 in the tens digit of the product. We have a 6 there already from the previous multiplication, and we can only add, not subtract, so we'll need to add a 9 --- the way to get a 9 is to multiply 13 by a 3

35 x 13 = 455

Now, we have a 4 in the hundreds place, so if we can add a one in that place we'd have a five. The way to 1 in the hundred's digit is to multiply 13 by 700

735 x 13 = 9555

Now, we have a 9 in the thousands place, so we need to add 6 --- multiply by 2000

2735 x 13 = 35555

Now, we have a 3 in the ten thousands place, so we need to add 2 --- multiply by 40000

42735 x 13 = 555555

Mirabilis dictu! As it happens, by windfall luck, we get not only a 5 in the ten thousands place, but also one in the hundred thousands place, making it a number with every digit equal to 5.

That's why D is the answer. Again, I can't imagine anyone in their right mind would expect you to slog through that without a calculator, and whatever else we say about the folks who write the GMAT, they are generally reasonable about this sort of thing.

Question #3:
3. The least number by which 72 must be multiplied in order to produce a multiple of 112, is:
a. 6
b. 12
c. 14
d. 18
e. none of the above

As I indicated above, this is bonafide, legitimate GMAT-type question. A questions like this do appear on the real GMAT.

The solution involves looking at the prime factorizations

72 = 8 * 9 = 2 * 2 * 2 * 3 * 3

112 = 2 * 56 = 2 * 8 * 7 = 2 * 2 * 2 * 2 * 7

So, the GCF of 72 and 112 is 2 * 2 * 2 = 8.

72 = 8 * 9

112 = 8 * 14

So, if we multiply 72 by 14, we will have a multiple of 112. In fact, in doing so, we will have created the LCM of 72 and 112. So the answer is C.

If those steps are unfamiliar, take a look at this blog article of mine: http://magoosh.com/gmat/2012/gmat-math-factors/

Question #4:
4. A number when divided by 3 leaves a remainder 1. When the quotient is divided by 2, it leaves a remainder1. What will be the remainder when the number is divided by 6

Again, I would say this is at the outer limit of what the GMAT might ask, assuming you were getting all the quant right and getting a steady diet of the hardest questions.

Same strategy as in Question #1 above -- work backwards.

Final quotient is q. To get that, we divided by 2 with a remainder of 1, so what we had before that division was 2q + 1.

That, in turn, was the quotient when we divided by 3 and get a remainder of 1, so what we had before that division was 3(2q + 1) + 1 = N

N = 3(2q + 1) + 1 = 6q + 4

Divide that by 6, and you get a remainder of 4.

====================================================================================================

Does all that make sense? If you have any questions, please do not hesitate to ask.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Re: Can Any one provide assistance??   [#permalink] 31 Jan 2012, 14:44
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