Leet remainder

Number Theory Level 1

What is the remainder when 1337 is divided by 101?


The answer is 24.

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Arron Kau by Arron Kau

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7 solutions

Sunil Pradhan
Dec 20, 2013

You can find using long division But using test of divisibility method

Test of divisibility of 101

Make groups of 2 digits from the end and add groups at odd places and add groups at even places.

find the difference. if difference is zero then the number is divisible by 101

Given number is 1337 Groups are 37 and 13 from the end

37 is the sum of groups at odd plaxes, abd 13 is sum of groups at even places. [in this example there is only one group]

37 – 13. = 24 shows the number is not divisible by 101 and remainder is 24

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Spelling mistakes : odd plaxes, abd 13 : read as odd places, and 13

Sunil Pradhan - 7 years, 5 months ago

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thank u

Delli babu - 7 years, 5 months ago

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Now consider 3713 here groups are interchanged. But carry the method as previous, always do subtraction of (sum of groups at odd places – sum of groups at even places) 13 – 37 = – 24 now remainder is not 24 but (101 – 24) = 77, same method is applicable to any test of divisibility using difference. e.g. in case of 11 also if difference is – ve then remainder is (11 – difference).

Sunil Pradhan - 7 years, 5 months ago

how 2 know d odd places & even places in 1337?

Delli babu - 7 years, 5 months ago

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start from right side, the first group is at odd place and 2nd at even, 3rd at odd 4th at even place and so on

the odd places group and even place may not be equal similarly there may not be 2 digits in the left most group. still consider it as a group. example of such is shown above.

similarly keep in mind it is not considered weather the digit is even or odd. Perhaps you might be knowing test for 11 where we consider group of 1 digit.

in 1337 the groups are 13 37 group of 37 (2 digits from the end is first group which is odd place and 13 (group of 2 digits) at the 2nd place which is even place..

Sunil Pradhan - 7 years, 5 months ago

odd places means 1st 3rd, ... and even means 2nd 4th, ... it is justlike test fot 11 sum of digits in alternate places, here there are 2 digits in each group.

e.g. 332290 gropus addition (90 + 33) odd places, and (22) is only group in even place e.g 2933227 odd palces (27 + 93) even places addition (32 + 2) subtraction is + 86 so remainder is 86

e,g. 2297210 sum (10 + 29) and (72 + 2) subtraction is – 35 so remainder is (101–35)= 66

Sunil Pradhan - 7 years, 5 months ago

is this test of divisibility method is different for every individual integer ?? for example.. if we have to divide any number 678942 from 143 and find the reminder then still can we use this divisibility method or if not then what method would you adopt ?

abhijeet pandey - 7 years, 5 months ago

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sorry for delay in reply as it was ovelooked. In case of 143 it is not prime 143 = 13 × 11

The combined test for 7, 11, 13 make groups of 3 digits from the end, add groups at odd places, add groups at even places, do subtraction 678 942 group at odd place is 942 and group at even place is 678

942 – 678 = 264

264 ÷ 143 remainder is 121

264 is divisible by 11 and 13 but not divisible by 7 to find remainder when 678942 ÷ 7 remainder is same as remainder of 264 ÷ 7 remainder is 26.

similarly think of, if number is 942678 then (678 – 942) = – 246 and actual remainder is (– 246) ÷ 143 = remainer is – 121 i.e (143 – 121) = 22

Sunil Pradhan - 7 years, 5 months ago

or you can use java! :P

A Former Brilliant Member - 7 years, 2 months ago

simple division

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Aashish Patel
Apr 14, 2014

it is simple.if you know division just divide and find the remainder.

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Java!!!!! :P

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Ria Patel
Dec 28, 2013

Divide 1337 by 101, and find the remainder.

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Jon Lin
Dec 21, 2013

1337/101 = 13.237...
101 x 13 = 1313
1337 - 1313 = 24

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1337/101 = 13.237
We can round this value up to the nearest tenth.
13.237 is approx. 13.24
24 is the remainder.
This method is viable for most if not all four digit numbers.



Jon Lin - 7 years, 5 months ago
Fizza Mubeen
Dec 19, 2013

1337/101=13.24 24 IS THE REMAINDER

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eh?

Gorou Hibiki - 7 years, 5 months ago

It seems it works for even 4 digit numbers. does not work for 5 digit numbers

Anant Pathak - 7 years, 5 months ago

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