Latest Elitmus Aptitude Question SOLUTION: Guyz i am unable yo understand remainder theorem..plz help me ..how to find remainder when 59^28 is dived by 7...similary 67^99 is divided 7...plz explain in detail...i need it...

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(#M40152442) ELITMUS Numerical Ability question remainder theorem Keep an EYE

Guyz i am unable yo understand remainder theorem..plz help me ..how to find remainder when 59^28 is dived by 7...similary 67^99 is divided 7...plz explain in detail...i need it...

Asked In Elitmus raj (8 years ago)

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raghavendra  (8 years ago)

67^99 can be written as (63+4)^99 which divided by 7gives 4^99....
4^99can be writtn as 4^3*33=64^33 and 64 can be written 63+1^33/7=1^33
1^33/7 so remainder is 1
according to me both question remainder is 1

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rachna  (8 years ago)

59^28 can be written as (56+3)^28 which divided by 7 gives 3^28 as remainder...
3^28 can be written as 27^9 *3 = (28-1)^9*3 which divided by 7 gives (-1)^9*3 as remainder.
so,(-1)^9*3=-3 can be written as (-7+4) which divided by 7 gives 4. so remainder is 4.

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Sonu kumar  (8 years ago)

you can be write 59^28 as (56+3)^28 ,56 is divisible by 7 then we can be write 3^28 as reminder...
3^28=now we can be write ((3^5)^5)*3^3.......3^5 =343 which is divisible by 7 then we can simply write..1^5*3^3=1*3*3*3=27..then we get reminder 6

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syed shadman akbar  (8 years ago)

Watch videos on YouTube by takshzilla shikshak first.. U won't feel any difficulty after that

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shyamala  (8 years ago)

in 59^28, take unit digit of 59 which is 9..
the repetitive squence of 9 is:
9^1=9
9^2=81
9^3=729 nd so on..
if yu observe the unit's digit in this sequence , it turns out to be 9,1,9,1,etc..(two numbers)

so divide 28 by 2.. remainder is 0
9 can be taken for odd remainders, 1 for even remainders..
since 0 is even we can take 1..
1 divided by 7.. remainder is 1.. ans is 1
hope you get me..

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prakash  (7 years ago)

according to me
ans is1

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Vikash kumar  (7 years ago)

please see the video of dinesh mirglani remainder chapter video yuour concept will be clear

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GAUTAM KUMAR  (7 years ago)

59^28
59 % 7 = 3
Hence the problem is converted into 3 ^ 28
7 ^ 1 =7
7 ^ 2 = 49
7 ^ 3 = 343
7 ^ 4 = 2301
7 ^ 5 = 16107
cycle is of 4
28 % 7 =0
Hence rem = 1.

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sarat  (7 years ago)

59^28/7 can be written as (56+3)/7 which is {(7n+x)/7 = x/7}; 3^28 /7.....deduce until (7n+1)/7=1 or x/7(x

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