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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
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To find the last two digits we can divide by 100 and find the remainder.
If we cancel out 4 from the denominator it becomes 25 and we can cancel out 4 from 440 in the numerator which then becomes 110.
Now we need to check for the remainder when divided by 25.
325 is completely divisible by 25 giving us a remainder of 0.
Only option possible is A.

Answer: A
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
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There is one multiple of 10, 2 and 5 so last two digits are 00.

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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
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Asked: What is the Last 2 digits of 238 X 234 X 198 X 237 X 199 X 325 X 674 X 876 X 440 X 139 X 217 X 719?
A) 00
B) 01
C) 37
D) 40
E) 58

Last 2 digits of 238 X 234 X 198 X 237 X 199 X 325 X 674 X 876 X 440 X 139 X 217 X 719
= Last 2 digits of 38*34*98*37*99*25*74*76*40*39*17*19; Since hundredth digits can not effect last 2 digits

Last digit = 0 ; Since 40 has 0
Second last digit of 38*34*98*37*99*25*74*76*40*39*17*19
= Last digit of 38*34*98*37*99*25*74*76*4*39*17*19 = 0 ; Since 25*4 = 100

Last 2 digits of 238 X 234 X 198 X 237 X 199 X 325 X 674 X 876 X 440 X 139 X 217 X 719 = 00

IMO A

Originally posted by Kinshook on 23 May 2020, 19:50.
Last edited by Kinshook on 23 May 2020, 20:55, edited 1 time in total.
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
IMO A

What is the Last 2 digits of 238 X 234 X 198 X 237 X 199 X 325 X 674 X 876 X 440 X 139 X 217 X 719?

238 X 234 X 198 X 237 X 199 X 325 X 674 X 876 X 440 X 139 X 217 X 719

325 = 25 X 13
440 = 4 X 10 X 11

325 X 440 = 4 X 10 X 11 x 25 X 13 = Last 2 digit 00

So product last 2 digit = 00
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
F(x) = 238 X 234 X 198 X 237 X 199 X 325 X 674 X 876 X 440 X 139 X 217 X 719

F(x) = 440 X (325X674) X K
These numerical terms give '00' at the end. Hence any multiplier 'K' will also give '00' at the end.
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
Last two digits will be 00

234 X 325 X 440
Will have 000 trailing behind them.
So no matter what. Rest of the calculation will have those zeroes trailing behind.

Answer is A

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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
multiply 440*325 and bang u will get three zero's at the end instead.
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
Answer : A

Aim : To see, if there are numbers that are either multiples of 5 & 2 or multiples of 10 to deduce if the last digits are 0

Fortunately, 440 (Multiple of 10) will contribute a 0 to the last two digits

325 along with one of the other even numbers will contribute another 0 to the last two digits.

Hence the last of 2 digits will be 00 (Option A)

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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
A

Take a hint from answer options , Options A & D end with 0.

So Try to isolate possible 10s from the stem.
440 -> Gives one 10
325 - > Multiplied with any even No. gives another 10

So, you have a Min of 2 Zeroes at the end.
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
325 * 674 will give you a multiple of 10 and 440 is again a multiple of 10.

So the last two digits will be 00.

Ans: A
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
Since 440 is a part of the list, we remark that the last digit of 238*....*219 (shorthand of the problem) is 0.
This leaves us with A and D, eliminating B,C and E. Now the last 2 digits of 325*440 can be roughly calculated as 25*40=1000.
Hence Option A is correct as the last 2 digits are 00.
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
You have at least two pairs of (2 * 5) prime factors that make up the large product

325 ——-> so final product will be divisible by 25 = (5)^2

More than 2 even numbers ———> so final product will be divisible by 4 = (2)^2

Since we have at least two pairs of (2 * 5) Prime Factors ———> we will have at least two terminating zeros

(A)
00

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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
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Re: What is the Last 2 digits of 238 * 234 * 198 * 237 * 199 * 325 * 674 * [#permalink]
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