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Compilation of tips and tricks to deal with remainders. : Quantitative Questions
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# Compilation of tips and tricks to deal with remainders.

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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

Ans: Using the technique: remainder = 0.12*t => the answer is multiple of 12. but none of the options match...did i miss something or is my understanding wrong
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ctrlaltdel wrote:
If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

Ans: Using the technique: remainder = 0.12*t => the answer is multiple of 12. but none of the options match...did i miss something or is my understanding wrong

Don't worry. This question is not that straightforward but understanding it will really help you for any other such question.

The answer is a multiple of 0.12.

However, If we consider it to be a multiple of 12, then we have to multiply each of the answers by 100 and then check.

Try working it out now.

You should get the answer to be (E). If you face any further difficulties then feel free to ask again!
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
sriharimurthy wrote:
ctrlaltdel wrote:
If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t?
(A) 2
(B) 4
(C) 8
(D) 20
(E) 45

Ans: Using the technique: remainder = 0.12*t => the answer is multiple of 12. but none of the options match...did i miss something or is my understanding wrong

Don't worry. This question is not that straightforward but understanding it will really help you for any other such question.

The answer is a multiple of 0.12.

However, If we consider it to be a multiple of 12, then we have to multiply each of the answers by 100 and then check.

Try working it out now.

You should get the answer to be (E). If you face any further difficulties then feel free to ask again!

Could you, please, explain the solution? I didn't get it.

I found out that the remaider is 3; should we just pick up the answer to get multiple of 3?
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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Quote:
Could you, please, explain the solution? I didn't get it.

I found out that the remaider is 3; should we just pick up the answer to get multiple of 3?

Hi Shelen,

In the question, it is given that s/t = 64.12

Thus, we know that the remainder in decimal format will be 0.12
(Note : Do not make the mistake of considering it to be 12. It is 0.12)

Now, we know that the answer should be a multiple of 0.12 since 'the remainder of s/t' will be equal to 'the remainder of s/t in decimal format' multiplied by 't'.

That is, R of (s/t) = 0.12*t ----> which is a multiple of 0.12 for all the positive integer values that 't' can hold.

Now, in order to make the calculation simpler, we can multiply both sides of the equation by 100.

R*100 = 12*t --> t = (R*100)/12

Now, since it is given that 't' is a positive integer, (R*100) has to be perfectly divisible by 12.

Thus look through the answer choices to see which one satisfies this condition.

You will find that R = 45 will be the only one that satisfies it, since 4500 is perfectly divisible by 12.

Therefore answer is choice (E) which is 45.
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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Wondering if it is a good way to solve it as follows:

.12 = 12/100 = 3/25 implies that the remainder is 3 or multiple of 3.

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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
brownybuddy wrote:
Wondering if it is a good way to solve it as follows:

.12 = 12/100 = 3/25 implies that the remainder is 3 or multiple of 3.

I tested it with some numbers. (8/5, 9/4, 57/12, 57/15, 57/20) They are work using the way you described. I don't know how to prove it mathematically though. Does anyone know? This is a smart way if it works for all numbers!
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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wilbase wrote:
sriharimurthy wrote:

Eg. $$R of \frac{(23)*(27)}{25} = R of \frac{(-2)*(2)}{24} = -4.$$ Now, since it is negative, we have to add it to 25.$$R = 25 + (-4) = 21$$

Is the "24" on the second part of the equation suppose to be "25"?

Yes. It is supposed to be 25. Thanks for spotting that. I have edited it.

brownybuddy wrote:
Wondering if it is a good way to solve it as follows:

.12 = 12/100 = 3/25 implies that the remainder is 3 or multiple of 3.

Yes. It follows from property number 7.

Since we are asked to find the remainder when 's' is divided by 't' and we are given the resulting number, we can write an equation as follows :

Remainder = (Decimal portion of the resulting number) * (Number we are dividing by)

Remainder = 0.12 * t

R = $$\frac{12}{100}*t$$ = $$\frac{3}{25}*t$$

So as you can see, the remainder 'R' must be a multiple of '3' provided 't' is an integer.

Since we know that 't' is an integer, we can safely conclude that 'R' is a multiple of '3'.

Note : In cases of remainder problems, even if 't' is not an integer it can be made into an integer. Eg. Remainder of 6/2.5 will be the same as Remainder of 12/5.
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
Hi ,

I am new on this forum , and i will tell you that what i have gained reading all the posts in the various sections have been mind blowing .

I wish to study indepth before i sit for my GMAT. I am quite ambitious with the kind of score i desire and i think you guys are the best in terms of detailing the requirements .

Please kindly explain the 3rd to 5th Rule on the remainder lecture ... i cant seem to grasp the rules!!

Thank you

Easy
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
Property 3 says that

Quote:
3) If a number has a remainder of ‘r’, all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of 21 is 5, then remainder of 7 (which is a factor of 21) will also be 5.

But if we see 21/5 remainder is 1
7 is a factor of 21
7/5 and the remainder is 2.

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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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msunny wrote:
Property 3 says that

Quote:
3) If a number has a remainder of ‘r’, all its factors will have the same remainder ‘r’ provided the value of ‘r’ is less than the value of the factor.
Eg. If remainder of 21 is 5, then remainder of 7 (which is a factor of 21) will also be 5.

But if we see 21/5 remainder is 1
7 is a factor of 21
7/5 and the remainder is 2.

Hi,

To make this example more clear : If any number when divided by 21 leaves a remainder of 5, then that number when divided by any factor of 21 will also leave a remainder of 5 provided the remainder is less than the factor.

Eg. R of 26/21 = 5

Factors of 21 are 3 and 7

Since 7 is greater than 5, R of 26/7 = 5

Since 3 is less than 5, R of 26/3 = R of 5/3 = 2

Hope this makes it clear.

I think I will edit the main post to make this point less confusing.

Cheers.
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
Hi ,

I am new on this forum , and i will tell you that what i have gained reading all the posts in the various sections have been mind blowing .

I wish to study indepth before i sit for my GMAT. I am quite ambitious with the kind of score i desire and i think you guys are the best in terms of detailing the requirements .

Please kindly explain the 3rd to 5th Rule on the remainder lecture ... i cant seem to grasp the rules!!

Thank you

Easy

Hi,

3rd Rule : I have explained the 3rd rule in the above post.

4th Rule : The cycle of powers is useful to know because it tells us the only possible values that the units place can hold for any particular number when it is raised to an integer power.

Go through the following example to see the usefulness of this rule :

Quote:
If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?
(1) n = 2
(2) m = 1

The Concept tested here is cycles of powers of 3.

The cycles of powers of 3 are : 3,9,7,1

St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.

St II) m=1 . This makes 3^(4*n +2 + 1).
4n can be 4,8,12,16...
3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFF
Hence B

5th Rule : Again for this rule, the best way to understand it is to work through a couple of questions (numbers-86325.html). Go through my solutions for the two problems in the post I have linked and see how rules 5 and 6 relate to them.

It might take a while for these concepts to get cemented but have a little patience and you will be rewarded.

If you have any specific doubts you would like me to address then please let me know.

Cheers.
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
Hello,

Can some one please tell me how to solve the below?

what is the remainder of 11^97/ 7

How do we tackle these type of problems?
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
Fokir wrote:
Hello,

Can some one please tell me how to solve the below?

what is the remainder of 11^97/ 7

How do we tackle these type of problems?

Let a mod b be the remainder when a is divided by b

Note that xy mod b = ((x mod b)*(y mod b)) mod b

11 mod 7 = 4
11^2 mod 7 = 4*4 mod 7 =2
11^3 mod 7 = 2*4 mod 7 =1
11^4 mod 7 = 1*4 mod 7 = 4
... and then the cycle will repeat

So for 11^97 the remainder will be 4
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
hi thanks a lot for these tips on remainders. Kudos to u.
I like the way u easily solved the 2 Remainder questions in some other posts. Could you also provide the links where you have solved questions involving the cycle of powers so as to get a better idea of that too.

thankssss
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
russty wrote:
hi thanks a lot for these tips on remainders. Kudos to u.
I like the way u easily solved the 2 Remainder questions in some other posts. Could you also provide the links where you have solved questions involving the cycle of powers so as to get a better idea of that too.

thankssss

Hi,

Here's a link regarding remainders and cycles of power. I found it very helpful. I guess this link was mentioned somewhere in this forum only.I had stored it in my favorite list..hope u too find it useful.

https://takshzilabeta.com/cat-quant/numb ... rs-part-i/

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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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An easy way to understand Number Tree:
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Re: Compilation of tips and tricks to deal with remainders. [#permalink]
great article. A suggestion - if you could add some examples of questions for each principle, it would be great for retention of the material. Thanks.
Re: Compilation of tips and tricks to deal with remainders. [#permalink]
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