Latest Elitmus Aptitude Question SOLUTION: Let R1 and R2 respectively denote the maximum and minimum possible remainders when (276)n is divided by 91 for any natural number n,n>= 144. Find R1+R2.

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(#M40157063) ELITMUS Numerical Ability question remainder theorem Keep an EYE

Let R1 and R2 respectively denote the maximum and minimum possible remainders when (276)n is divided by 91 for any natural number n,n>= 144. Find R1+R2.

Asked In Elitmus Shivansh Shrivastava (8 years ago)

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Subhabrata Sarkar  (8 years ago)

But the least value for n can be= 144 (as given in question, n>=144)

So, 3n/91= 3*144/91= 68

now, the highest remainder can be when it reaches 68+3+3+3+3+3+3+3= 89

(as each time we increment n by 1 the remainder increases by 3,
3*145/91= 71- remiander
3*146/91= 74
3*147/91=77
... so on and so far upto 89 after which it becomes 89+3= 92 which means it crosses 91)

Hence least remainder would be 92 -91=1
and highest remainder = 89
R1+ R2= 89+1 = 90

Is this solution Helpfull? Yes (20) | No (4)  

samiksha  (8 years ago)

The remainder when 276 divided by 91 is 3 hence

remainder of (276)n/91 = remainder 3n/91

Taking n = 1,2,3,4,5

We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.

For n>5, the pattern repeats

Hence R1 = 81 and R2 = 1

R1+R2 = 81+1 = 82

Is this solution Helpfull? Yes (5) | No (6)  

monika  (8 years ago)

The remainder when 276 divided by 91 is 3 hence

remainder of (276)n/91 = remainder 3n/91

Taking n = 1,2,3,4,5

We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.

For n>5, the pattern repeats

Hence R1 = 81 and R2 = 1

R1+R2 = 81+1 = 82

Is this solution Helpfull? Yes (2) | No   

satish  (8 years ago)

any number say m when divide another say n . contain minimum remainder 0 and maximum m-1.
so in case of (276)n / 91 . no need to calculate answer will be R1+R2= 0+(91-1)= 90

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Aditya  (8 years ago)

@himanshu send me study material for preparing elitmus n my email id- adikumar1995@gmail.com

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subhajit chakraborty  (8 years ago)

ans would be 91..the max remainder can be 90n the minimum remainder can be 1 so R1+R2 would be.(90+1=91)

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subhajit chakraborty  (8 years ago)

sorry in my earlier solution i did a mistake ...276*n/91 the remainder will be 3*n/91,now if i devide n by 91 the maximum remainde can be 90 so 3*90=270 gain devide it by 91 the remainder would be 88..
now in the menimum case n/91 the minimum remainder can be 1 so3*1/91 rem is 3 hence r1=3,r2=87 so total sum would be=87+3=91

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Madhu  (7 years ago)

he remainder when 276 divided by 91 is 3 hence

remainder of (276)n/91 = remainder 3n/91

Taking n = 1,2,3,4,5

We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.

For n>5, the pattern repeats

Hence R1 = 81 and R2 = 1

R1+R2 = 81+1 = 82

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Sayantani Maiti  (7 years ago)

The remainder when 276 divided by 91 is 3 hence

remainder of (276)n/91 = remainder 3n/91

Taking n = 1,2,3,4,5

We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.

For n>5, the pattern repeats

Hence R1 = 81 and R2 = 1

R1+R2 = 81+1 = 82

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Suvankar Saha  (5 years ago)

The remainder when 276 divided by 91 is 3 hence,
remainder of (276)n/ 91 = remainder 3n/91
Taking n = 1,2,3,4,5.
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats.
Hence, R1 = 81 and R2 = 1.
and, R1 + R2 = 81 + 1 = 82.

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