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What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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What is the remainder when $$43717^{(43628232)}$$ is divided by 5

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Originally posted by guerrero25 on 25 Jun 2013, 15:10.
Last edited by Bunuel on 12 May 2017, 03:08, edited 3 times in total.
Edited the question.
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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10
5
mikemcgarry wrote:
guerrero25 wrote:
What is the remainder when $$[(((((((43717)^4)^3)^6)^2)^8)^2)^3]^2$$ is divided by 5

(A)1
(B)2
(C)3
(D)4
(3)5

Dear guerrero25
First of all, I believe all the parentheses I added are absolutely necessary to make clear what the question is asking. Remember that parentheses are not mathematical garnish, like parsley served with a meal. They have a crucial function in many problems.

The problem is way-over-the-top harder than anything the GMAT would ask. Having said that, the principle is something the GMAT does test. The principle is: in large powers, the units digit of the power depends only on the units digit of the base. More generally, if we multiply A x B = C, the units digit of C is determined exclusively by the units digits of A & B --- none of the other digits of A & B have any influence on the units digit of C.

First of all, we can totally ignore the first four digits of the base, 43717 ---- only the 7 at the end matters.
What is the units digit of 7^4?
7x7 = 49, so that's a units digit of 9.
9 x 7 = 63, so (7^3) has a units digit of 3
3 x 7 = 21, so (7^4) has a units digit of 1

Well, this is an incredible stroke of luck, because we now have a units digit of 1, and all subsequent powers of this will be 1, because 1 to any power is simply 1. Therefore, the final power, some god-awful number, must have a units digit of 1, and when divided by 5, it has a remainder of 1. Answer = (A).

BTW, just out of curiosity, I checked this number on Wolfram Alpha. The power stated in the prompt, if fully calculated out, would have 64,153 decimal pages. In typical fonts (around 3000 characters per page), it would take 21 full pages and would spill onto the 22nd page to print this number out. A large-ish number!

Mike Original question reads: What is the remainder when 43717^(43628232) is divided by 5?

The remainder when 43717^(43628232) is divide by 5 will be the same as the remainder when 7^(43628232) is divided by 5 (we need only the units digit to get the remainder upon division by 5).

7^1=7 divided by 5 yields the remainder of 2;
7^2=49 divided by 5 yields the remainder of 4;
7^3=343 divided by 5 yields the remainder of 3;
7^4=...1 divided by 5 yields the remainder of 1.
7^5=...7 divided by 5 yields the remainder of 2 AGAIN.

The remainders repeat in blocks of four {2, 4, 3, 1}, {2, 4, 3, 1}, ...

43628232 (exponent) is divisible by 4 (a number is divisible by 4 if its last 2 digits (32 in our case) divisible by 4). Therefore, the remainder when 43717^(43628232) is divided by 5 is the fourth number in pattern, which is 1.

Hope it's clear.
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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6
5
guerrero25 wrote:
What is the remainder when $$[(((((((43717)^4)^3)^6)^2)^8)^2)^3]^2$$ is divided by 5

(A)1
(B)2
(C)3
(D)4
(3)5

Dear guerrero25
First of all, I believe all the parentheses I added are absolutely necessary to make clear what the question is asking. Remember that parentheses are not mathematical garnish, like parsley served with a meal. They have a crucial function in many problems.

The problem is way-over-the-top harder than anything the GMAT would ask. Having said that, the principle is something the GMAT does test. The principle is: in large powers, the units digit of the power depends only on the units digit of the base. More generally, if we multiply A x B = C, the units digit of C is determined exclusively by the units digits of A & B --- none of the other digits of A & B have any influence on the units digit of C.

First of all, we can totally ignore the first four digits of the base, 43717 ---- only the 7 at the end matters.
What is the units digit of 7^4?
7x7 = 49, so that's a units digit of 9.
9 x 7 = 63, so (7^3) has a units digit of 3
3 x 7 = 21, so (7^4) has a units digit of 1

Well, this is an incredible stroke of luck, because we now have a units digit of 1, and all subsequent powers of this will be 1, because 1 to any power is simply 1. Therefore, the final power, some god-awful number, must have a units digit of 1, and when divided by 5, it has a remainder of 1. Answer = (A).

BTW, just out of curiosity, I checked this number on Wolfram Alpha. The power stated in the prompt, if fully calculated out, would have 64,153 decimal pages. In typical fonts (around 3000 characters per page), it would take 21 full pages and would spill onto the 22nd page to print this number out. A large-ish number!

Mike _________________
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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3
Original question reads: What is the remainder when 43717^(43628232) is divided by 5?

The remainder when 43717^(43628232) is divide by 5 will be the same as the remainder when 7^(43628232) is divided by 5 (we need only the units digit to get the remainder upon division by 5).

7^1=7 divided by 5 yields the remainder of 2;
7^2=49 divided by 5 yields the remainder of 4;
7^3=343 divided by 5 yields the remainder of 3;
7^4=...1 divided by 5 yields the remainder of 1.
7^5=...7 divided by 5 yields the remainder of 2 AGAIN.

The remainders repeat in blocks of four {2, 4, 3, 1}, {2, 4, 3, 1}, ...

43628232 (exponent) is divisible by 4 (a number is divisible by 4 if its last 2 digits (32 in our case) divisible by 4). Therefore, the remainder when 43717^(43628232) is divided by 5 is the fourth number in pattern, which is 1.

Hope it's clear.[/quote]

Hi Bunuel,

The exponent 43628232 is an even no and hence 7^(even no) will end up either in 9 or 1 and therefore we can have remainder on dividing by 5 as 4 or 1 respectively.
How did you chose A as because when 43628232 is divisible by 2 and 4 both..

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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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mridulparashar1 wrote:
Hi Bunuel,

The exponent 43628232 is an even no and hence 7^(even no) will end up either in 9 or 1 and therefore we can have remainder on dividing by 5 as 4 or 1 respectively.
How did you chose A as because when 43628232 is divisible by 2 and 4 both..

I thought that was explained....

The remainders repeat in blocks of four {2, 4, 3, 1}, {2, 4, 3, 1}, ... 43628232 (exponent) is divisible by 4, thus the remainder will be 4th number in pattern.
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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Bunuel wrote:
mridulparashar1 wrote:
Hi Bunuel,

The exponent 43628232 is an even no and hence 7^(even no) will end up either in 9 or 1 and therefore we can have remainder on dividing by 5 as 4 or 1 respectively.
How did you chose A as because when 43628232 is divisible by 2 and 4 both..

I thought that was explained....

The remainders repeat in blocks of four {2, 4, 3, 1}, {2, 4, 3, 1}, ... 43628232 (exponent) is divisible by 4, thus the remainder will be 4th number in pattern.

Yes...you did.

I missed the part of dividing the exponent with cyclicity of 7 which is 4 as you mentioned.
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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Dear Brunel ,

In case exponent is not completely divisible by 4 ..lets say it leaves a remainder of 2 ..then what will be answer .
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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Bunuel wrote:
mikemcgarry wrote:
guerrero25 wrote:
What is the remainder when $$[(((((((43717)^4)^3)^6)^2)^8)^2)^3]^2$$ is divided by 5

(A)1
(B)2
(C)3
(D)4
(3)5

Dear guerrero25
First of all, I believe all the parentheses I added are absolutely necessary to make clear what the question is asking. Remember that parentheses are not mathematical garnish, like parsley served with a meal. They have a crucial function in many problems.

The problem is way-over-the-top harder than anything the GMAT would ask. Having said that, the principle is something the GMAT does test. The principle is: in large powers, the units digit of the power depends only on the units digit of the base. More generally, if we multiply A x B = C, the units digit of C is determined exclusively by the units digits of A & B --- none of the other digits of A & B have any influence on the units digit of C.

First of all, we can totally ignore the first four digits of the base, 43717 ---- only the 7 at the end matters.
What is the units digit of 7^4?
7x7 = 49, so that's a units digit of 9.
9 x 7 = 63, so (7^3) has a units digit of 3
3 x 7 = 21, so (7^4) has a units digit of 1

Well, this is an incredible stroke of luck, because we now have a units digit of 1, and all subsequent powers of this will be 1, because 1 to any power is simply 1. Therefore, the final power, some god-awful number, must have a units digit of 1, and when divided by 5, it has a remainder of 1. Answer = (A).

BTW, just out of curiosity, I checked this number on Wolfram Alpha. The power stated in the prompt, if fully calculated out, would have 64,153 decimal pages. In typical fonts (around 3000 characters per page), it would take 21 full pages and would spill onto the 22nd page to print this number out. A large-ish number!

Mike Original question reads: What is the remainder when 43717^(43628232) is divided by 5?

The remainder when 43717^(43628232) is divide by 5 will be the same as the remainder when 7^(43628232) is divided by 5 (we need only the units digit to get the remainder upon division by 5).

7^1=7 divided by 5 yields the remainder of 2;
7^2=49 divided by 5 yields the remainder of 4;
7^3=343 divided by 5 yields the remainder of 3;
7^4=...1 divided by 5 yields the remainder of 1.
7^5=...7 divided by 5 yields the remainder of 2 AGAIN.

The remainders repeat in blocks of four {2, 4, 3, 1}, {2, 4, 3, 1}, ...

43628232 (exponent) is divisible by 4 (a number is divisible by 4 if its last 2 digits (32 in our case) divisible by 4). Therefore, the remainder when 43717^(43628232) is divided by 5 is the fourth number in pattern, which is 1.

Hope it's clear.

In which cases can we rely on just picking the units digit like in this case? Is it applicable to all single digit divisors apart from 5?

Cheers!
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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guerrero25 wrote:
What is the remainder when 43717^(43628232) is divided by 5

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Last digit of powers of 7
7^1 - 7
7^2 - 9
7^3- 3
7^4 - 1
7^5 - 7 and the process repeats again, which means every power which is divisible by 4 will have last digit as 1.

43628232 is divisible by 4 - check the last two digits

Divide this number by 5 the remainder will be 1 as the unit digit is 1
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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kanikaamar wrote:
is there any other simpler way to answer this question??

When 43717 is divided by 5, the remainder is 2. Now 2^(43628232)/5 must be calculated. The term can be rewritten as 4^21814116/5.
4/5 leaves remainder -1; hence the final remainder is (-1)^21814116 =1
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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$$\frac{43717^{43628232}}{5}$$

Last 3 digits are divisible by 4;so for the calculation of remainder, result would be the same

$$7^1 = 7$$

$$7^2 = 49$$

$$7^3 = 343$$

$$7^4 = 2401$$

Remainder for $$\frac{43717^{43628232}}{5}$$ would be 1

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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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1
The last digits of power of any number follow a pattern.

7^1 = 7
7^2 = 49

the last digit of (7^2) is '9'. so the unit place of next power of 7 will be 3 ( 9*7 = 63, unit digit is 3) and similarly the next power of 7 will end with '1'

so the digits, 7,9,3,1 repeat. If the power is divisible by 4, thn the unit digit will be 1. If 43628232 is divisible by 4, so the unit digit of '43717^43628232' is 1.

And when divided by 5, the reminder is 1.
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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guerrero25 wrote:
What is the remainder when 43717^(43628232) is divided by 5

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

We can use the modulo theorem here
Remainder when $$43717^{43628232}$$ divided by $$5$$ will be same as when $$mod(43717,5)^{43628232}$$
so $$43717 mod 5 = 2$$ and $$2$$ repeats every $$4th$$ powers as with last digit as $$(2,4,8,6)$$
and $$(43628232)$$ does gets divided by $$4$$. so the last digit will be $$6$$ and hence remainder when divided by $$5$$ is 1. A
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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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guerrero25 wrote:
What is the remainder when $$43717^{(43628232)}$$ is divided by 5

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

To determine the remainder when 43717^(43628232) is divided by 5, we need the units digit of that number. Thus, we really care about the units digit of 7^(43628232).

Let’s evaluate the pattern of the units digits of 7^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are ONLY concerned with the units digit of 7 raised to each power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digit of powers of 7 repeats every 4 exponents. The pattern is 7–9–3–1. In this pattern, all positive exponents that are multiples of 4 will produce 1 as its units digit.

Since the last two digits of 43628232 are 32, which is divisible by 4, we know that 43628232 is a multiple of 4. Thus, 7^(43628232) has a units digit of 1, and thus it will leave a remainder of 1 when divided by 5.

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Re: What is the remainder when 43717^(43628232)is divided by 5  [#permalink]

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Is this type of question ever asked in Real GMAT?
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_________________ Re: What is the remainder when 43717^(43628232)is divided by 5   [#permalink] 04 Oct 2019, 08:40
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