Find the last 2 digits of 65*29*37*63*71*87*62

Yes. You cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, remember, if you do want to cancel off to make life easier for you, you can do it, provided you remember to multiply it back again.
So say, I want to find the remainder when 14 is divided by 10 (remainder 4) and cancel off the common 2 to get 7 divided by 5 giving me a remainder of 2, I can multiply back the 2 I canceled with the remainder to get a remainder of 4. That's a valid technique.

e.g. What is the remainder when 85 is divided by 20? It is 5.
or I might rephrase it as what is the remainder when 17 is divided by 4 (I cancel off 5 from the numerator and the denominator). The remainder in this case is 1. I multiply the 5 back to 1. I get the remainder as 5!

Can someone explain the highlighted part. Why R of (x-n)/n = R of (6-25)/25 = +6?

Q1.

For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows.

5*-1*-3*3*1*-3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90.

Q2.

same approach. 1*2*3*4*-4*-3-*-2-*-1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36.

To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36

I am not sure from where you are reading this and what this means but I think you are confused with the concept of negative remainders. It is discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... emainders/

When the remainder of x/n is 6, there is no confusion. But when the remainder of x/n comes out to be -6, we need to adjust the remainder (make it positive) by making it n - 6 (why has been discussed in the post)

Also, when talking about divisibility and remainders, you don't usually take negative integers into account. Divisibility is a positive integer concept. You can divided 6 balls among 3 kids but not -6 balls among 3 kids.

So (6-25) = -19 divided by 25 doesn't really make sense.

If instead it were (25n - 19) divided by 25, then the remainder will be -19 since 25n is divisible by 25. A remainder of -19 is same as a remainder of 6.

Hi Karishma, many thanks for your response. I am a fan of your blogs, do see my Quant Consolidated notes, Carcass's efforts of your blogs are best in there...

My confusion is how do we get to remainder 6 and not -19?
6/25 if this has remainder 6 then why should 24/25 have remainder of -1 (is the formula not same (x-n)/n
The negative part I understand completely, your blogs are too good

When the divisor is 25, a remainder of 6 is the same thing as a remainder of -19. Since GMAT doesn't give you negative remainders, you will not have -19 in the options. So you mush choose 6.
24/25 has remainder 24 which is also same as remainder of -1. Both are correct but again, GMAT will only give you 24 in the options.
Think of it: when you have 24 balls and you must distribute them equally among 25 kids, you can say that you have 24 balls remaining and you gave each kid 0 (the quotient) ball. Or you can say that you have -1 ball remaining (i.e. you gave one extra ball from your side) and each kid got 1 (the quotient) ball.

Q1.
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

Finding the last two digits of an expression


Thru the Application of REMAINDER THEOREM


Remainder of the above expression when divided by 100 will give the answer of this question

\(\frac{65*29*37*63*71*87*62}{100}\)

\(\frac{13*29*37*63*71*87*62}{20}\) ------------------> on dividing by 5

\(\frac{13*9*17*3*11*7*2}{20}\) ------------------> taking remainder when each number divided by 20

\(\frac{117*51*77*2}{20}\)

\(\frac{17*11*17*2}{20}\) ------------------> taking remainder when each number divided by 20

\(\frac{289*22}{20}\)

\(\frac{9*2}{20}\) ------------------> taking remainder when each number divided by 20

\(\frac{18}{20}\)

So 18 is the remainder
However since initially we divided the numerator and denominator by 5, now we need to multiply 18 by 5

So last two digits = 90

Thru the Application of NEGATIVE REMAINDERS


\(\frac{13*29*37*63*71*87*62}{20}\)

\(\frac{(-7)*9*(-3)*3*(-9)*7*2}{20}\) ----------> taking remainder when each number divided by 20

\(\frac{21*27*(-9)*14}{20}\)

\(\frac{1*7*(-9)*(-6)}{20}\) ----------> taking remainder when each number divided by 20

\(\frac{7*54}{20}\)

\(\frac{(-13)*(-6)}{20}\) ----------> taking remainder when each number divided by 20

\(\frac{78}{20}\)

\(\frac{18}{20}\) ----------> taking remainder when each number divided by 20

18*5 = 90

NOTE :- If we are asked to find last 3 digits of the expression, we will obtain individual remainders when divided by 1000

Although Indian CAT is fond of such problems, i have never seen those in my any GMAT practice material

Regards,

Narenn

49=100-51 but not 41. Or it doesn't matter?

Could you also explain how to calculate that (100^2 - 44^2) has 64 as last two digits?




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