An integer N is selected at random in the range 1 N 2020. What is

Question

An integer N is selected at random in the range 1 ≤ N ≤ 2020. What is the probability that the remainder when N^16 is divided by 5 is 1?

(A) \frac{1}{5}

(B) \frac{2}{5}

(C)  \frac{3}{5}

(D) \frac{4}{5}

(E) 1

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Nidzo · Accepted Answer

For a number to have a remainder of 1 when divided by 5, the units digit of the number must be either 1 or 6.

The easiest approach is to view the exponent patterns of 0 through to 9, and see if when they are raised to the power of 16 if they end in 1 or 6.

0 and 5 can immediately be discarded as any number with 0 or 5 as their units digit will be divisible by 5.

1-  Any number with 1 as it's unit digit, raised to any power will have 1 as its unit digit.  Will leave a remainder of 1 when divided by 5

2-   2^{16}  will have a units digit of 6.  Will leave a remainder of 1 when divided by 5

3-   3^{16}  will have a units digit of 1.  Will leave a remainder of 1 when divided by 5

4-   4^{16}  will have a units digit of 6.  Will leave a remainder of 1 when divided by 5

6-   6^{16}  will have a units digit of 6.  Will leave a remainder of 1 when divided by 5

7-   7^{16}  will have a units digit of 1.  Will leave a remainder of 1 when divided by 5

8-   8^{16}  will have a units digit of 6.  Will leave a remainder of 1 when divided by 5

9-   9^{16}  will have a units digit of 1.  Will leave a remainder of 1 when divided by 5

This means that 8 from the 10 possible unit digits when raised to the power of 16 will leave a remainder of 1 when divided by 5.

Probability that  \frac{n^{16}}{5}  leaves a remainder of 1:  \frac{8}{10}  or simplified  \frac{4}{5}

Answer D

RiyaJain69 · Answer

What a question! And what a solution! I have been struggling in the quant section. I have learnt all these concepts in the GMAT course (Wizako) I did but questions such as these are so tricky and new to me that I'm finding myself like a deer caught in headlights, unable to proceed. I feel lost. This makes me nervous and prone to choking in further questions! Has anybody got any suggestions for me?

ThatDudeKnows · Answer

RiyaJain69

If you get a question of this difficulty on the real test, you should do the absolute opposite of choking, as it means you are well on your way to a Q49+. Keep in mind that if you're shooting for something like Q49, the algorithm expects you to miss HALF of the questions that are of that level or harder. So, if you can spot a trick on the other half, you're in good shape.

ThatDudeKnows · Answer

In order to have a remainder of 1 when divided by 5, we need a units digit of 1 or 6; none of the other digits matter.

When any number is raised to increasing powers, the units digits cycle as follows:

If the original number ends in 1: 1, 1, 1, 1 repeat
If the original number ends in 2: 2, 4, 8, 6 repeat
... 3: 3, 9, 7, 1 repeat
4: 4, 6, 4, 6 repeat
5, 5, 5, 5, 5 repeat
6: 6, 6, 6, 6 repeat
7: 7, 9, 3, 1 repeat
8: 8, 4, 2, 6 repeat
9: 9, 1, 9, 1 repeat
0: 0, 0, 0, 0 repeat

If we raise anything to the 2020th power, we complete 505 of those cycles. Eight of them will be on a units digit that ends in a 1 or 6.

8/10 = 4/5

Answer choice D.

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=BmCcQyCfB2I

Given that An integer N is selected at random in the range 1 ≤ N ≤ 2020. What is the probability that the remainder when N^16 is divided by 5 is 1?

Theory : To find the remainder of a number by 5, we can just find the remainder of the unit's digit of the number by 5 
(Watch  this video  to  Learn the Divisibility Rules )

Since we need to get a remainder of 1 after dividing by 5 => the unit's digit can be
0 + 1 and 5 + 1 = 1 and 6 respectively

Let's look at all the numbers’ from 1 to 10 and see whose  16^{th}  power can give us 1 or 6 as the unit's digit

Unit's digit of any power of 1 will always be 1 => 1^16 will give 1 remainder when divided by 1 =>   POSSIBLE  
=>  16^{th}  power of any number with 1 as the unit's digit will give 1 remainder when divided by 5

Unit's digit of power of 2 will be 2, 4, 8, 6, 2, 4.... => They repeat after 4 => Unit's digit of 2^16 = unit's digit of 2^4 = 6 =>   POSSIBLE

Unit's digit of power of 3 will be 3, 9, 7, 1, 3, 9.... => They repeat after 4 => Unit's digit of 3^16 = unit's digit of 3^4 = 1 =>   POSSIBLE

Unit's digit of power of 4 will be 4, 6, 4, 6, 4, 6.... => They repeat after 2 => Unit's digit of 4^16 = unit's digit of 4^4 = 6 =>   POSSIBLE

Unit's digit of all powers of 5 will be 5 itself =>   NOT POSSIBLE

Unit's digit of all powers of 6 will be 6 itself =>   POSSIBLE

Unit's digit of power of 7 will be 7, 9, 3, 1, 7,... => They repeat after 4 => Unit's digit of 7^16 = unit's digit of 7^4 = 1 =>   POSSIBLE

Unit's digit of power of 8 will be 8, 4, 2, 6, 8,... => They repeat after 4 => Unit's digit of 8^16 = unit's digit of 8^4 = 6 =>   POSSIBLE

Unit's digit of power of 9 will be 9, 1, 9, 1, 9.... => They repeat after 2 => Unit's digit of 9^16 = unit's digit of 9^4 = 1 =>   POSSIBLE

Unit's digit of all powers of 10 will be 0 itself =>   NOT POSSIBLE

=> All numbers with unit's digit as 1, 2, 3, 4, 6, 7, 8, 9 from 1 to 2020 will give 1 remainder when divided by 5
=> 1, 2, 3, 4, 6, 7, 8, 9 are 8 numbers out of 10 numbers from 1 to 10, similar ratio will be there from 1 to 2020

=> Probability(N divided by 5 will give remainder of 1) =  \frac{8}{10}  =  \frac{4}{5}

So,  Answer will be D 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

jadechan · Answer

To find the probability that the remainder when  N^{16}  is divided by 5 is 1, we can consider the possible values of N in terms of multiples of 5.

Let's express N as: 
 N = 5k, 5k+1, 5k+2, 5k+3, 5k+4

Since  5k+3 = 5k-2  and  5k+4 = 5k-1 , we can rewrite the expressions as: 
 N = 5k, 5k+1, 5k+2, 5k-2, 5k-1

Now, let's evaluate  N^{16}  for each case:

(1)  (5k)^{16} = 5^{16}k^{16}  = 0 (when divided by 5)

(2)  (5k+1)^{16} = 5k'+1  = 1 (when divided by 5)

(3)  (5k+2)^{16} = ((5k+2)^4)^4 = (5k+2^4)^4 = 5k'+(2^4)^4 = 5k'+1  = 1 (when divided by 5)

(4)  (5k-2)^{16} = ((5k-2)^4)^4 = (5k+(-2)^4)^4 = 5k'+(-2^4)^4 = 5k'+1  = 1 (when divided by 5)

(5)  (5k-1)^{16} = ((5k-1)^4)^4 = (5k+(-1)^4)^4 = 5k'+(-1)^{16} = 5k'+1  = 1 (when divided by 5)

Out of these 5 cases, 4 of them result in a remainder of 1 when divided by 5. Therefore, the probability that the remainder when  N^{16}  is divided by 5 is 1 is:

Probability =  \frac{4}{5}

So, the correct answer is (D)  \frac{4}{5} .

HarshavardhanR · Answer

Given :  N is selected at random in the range 1 ≤ N ≤ 2020.
  Asked : P robability that the remainder when N^16 is divided by 5 is 1

-> For a number to give a remainder of 1 when divided by 5,  the unit's digit of the number has to be either 1 or 6 .

-> Total number of numbers -> 1^16, 2^16, 3^16..........2020^16 -> a total of  2020  numbers. (1)

-> How many of the above 2020 numbers have unit digit 1?

(1) Any number ending with 1^anything = 1.

1^16, 11^16, 21^16,........2011^16 =  \frac{(2011 - 1) }{ 10}  + 1 = 202

(2) Numbers ending with 3, when raised to the 16th power also yield unit digit 1 (3-9-7- 1, 4th power gives 1 ).

3^16, 13^16, 23^16,............................2013^16 =  \frac{(2013 - 3) }{ 10}  + 1 = 202

(3) Numbers ending with 7, when raised to the 16th power also yield unit digit 1 (7-9-3- 1, 4th power gives 1 ).

7^16, 17^16, 27^16,........2017^16 =  \frac{(2017 - 7) }{ 10}  + 1 = 202

(4) Numbers ending with 9, when raised to even power also yield unit digit 1 (9-1-9-1, so  16th power gives 1 ).

9^16, 19^6,.................2019^16 =  \frac{(2019 - 9) }{ 10}  + 1 = 202

-> How many of the above 2020 numbers have unit digit 6?

(1) Any number ending with 6^anything = 6.

6^16, 16^16, 26^16,........2016^16 =  \frac{(2016 - 6) }{ 10}  + 1 = 202

(2) Numbers ending with 2, when raised to the 16th power also yield unit digit 1 (2-4-8-6 , 4th power gives 6 ).

2^16, 12^16, 22^16,............................2012^16 =  \frac{(2012 - 2) }{ 10}  + 1 = 202

(3) Numbers ending with 8, when raised to the 16th power also yield unit digit 1 (8-4-2-6 , 4th power gives 6 ).

8^16, 18^16, 28^16,........2018^16 =  \frac{(2018 - 8) }{ 10}  + 1 = 202

(4) Numbers ending with 4, when raised to even power also yield unit digit 1 (4-6-4-6, so  16th power gives 6 ).

4^16, 14^6,.................2014^16 =  \frac{(2014 - 4) }{ 10}  + 1 = 202

So, total number of numbers which will give a remainder of 1 when divided by 5 =  8 x 202

Probability( remainder when N^16 is divided by 5 is 1) =  \frac{Favorable outcomes }{ total outcomes } =  \frac{8 * 202 }{ 2020}  =  \frac{8}{10}  =  \frac{4}{5}  (Choice E).

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Harsha

An integer N is selected at random in the range 1 N 2020. What is

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