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

Question

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

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

Bunuel · Accepted Answer

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.

Answer: A.

Hope it's clear.

mikemcgarry · Answer

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 :-)

WoundedTiger · Answer

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.

Answer: A.

Hope it's clear.

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..

Please elaborate.

Bunuel · Answer

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.

WoundedTiger · Answer

Yes...you did.

I missed the part of dividing the exponent with cyclicity of 7 which is 4 as you mentioned.

archit · Answer

Dear Brunel ,

In case exponent is not completely divisible by 4 ..lets say it leaves a remainder of 2 ..then what will be answer .

Bunuel · Answer

If the remainder when the exponent divided by 4 where 2, then the remainder when 43717^(exponent) divided by 5 would be the second number in the pattern, thus 4. Similar question to practice: what-is-the-remainder-when-43-86-is-divided-by-134778.html when-51-25-is-divided-by-13-the-remainder-obtained-is-130220.html what-is-the-remainder-of-126493.html what-is-the-remainder-when-32-32-32-is-divided-by-100316.html what-is-the-remainder-when-18-22-10-is-divided-by-99724.html what-is-the-remainder-when-333-222-is-divided-by-156379.html to-find-the-units-digit-of-a-large-number-155363.html if-x-is-a-positive-integer-is-the-remainder-0-when-3-x-109075.html if-x-and-y-are-positive-integers-what-is-the-remainder-when-109636.html if-you-divide-7-131-by-5-which-remainder-do-you-get-83350.html if-r-s-and-t-are-all-positive-integers-what-is-the-136746.html what-is-the-remainder-when-3-243-is-divided-by-141050.html Theory on remainders problems: remainders-144665.html DS remainders problems to practice: search.php?search_id=tag&tag_id=198 PS remainders problems to practice: search.php?search_id=tag&tag_id=199 Hope it helps.

jlgdr · Answer

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! J

PerfectScores · Answer

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

Amit0507 · Answer

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

PareshGmat · Answer

\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

Answer = 1 = A

sshrivats · Answer

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.

Senthil1981 · Answer

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

JeffTargetTestPrep · Answer

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.

Answer: A

Shri15kumar · Answer

Is this type of question ever asked in Real GMAT?

Basshead · Answer

Cyclicity of 7 = 7, 9, 3, 1.

43628232 is divisible by 4 because the last two digits (32) is divisible by four. Therefore, the units digit of  43717^{43628232}  is 1.

1 divided by 5 gives us remainder 1.

Answer is A.

Deyoz · Answer

can some one please tell whether my approach to solve the problem is correct?

43717^(43628232)
just considering only 7^ (43628232)

= 7^2(21814116)
=49^(21814116)
={(50-1)^(21814116)}/5
=r{(0 -1)^(21814116)}/5
=r -1 raised to the even power equals 1 ....so the choice A is correct

bv8562 · Answer

Hey  Bunuel , could you please confirm if I have solved the question correctly?

\frac{43717^{43628232}}{5}  =  \frac{(43720-3)^{43628232}}{5}  =  \frac{(-3)^{43628232}}{5}  =  \frac{(-3)^4}{5}  =  \frac{81}{5}  => Remainder = 1  (A)

brains · Answer

Took me 27 secs.
Any time you see bigger number , try to break down sum of two number,one of which is divisible by divisor. Here we can break into 43715+2.
Now the exponent 43628232, which is divisible by 4. So we can safely assume that the cyclicity will be 4, since the extra number is 2 in the sum. Remember when you expand the expression, you will get only one constant term independent of 43715, and that will be (2^)43628232. All the other terms will have 43715 which is divisible by 5.

Since the divisor is 5, the question remains what is the remainder when 2^(43628232) is divisible by 5. SInce the the exponent is divisible by 4 and the cyclicity is 4 for 2, we have the unit digit as 6. So the remainder will be 1.

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

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

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