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Numerical Ability
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What will be the reminder when (1234567890123456789)^24 is divided by 6561 ?
Read Solution (Total 6)
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- The digital sum of number is
(1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9)=90=(9+0)=9
So it is divisible by 3 or 9
The given divisor 6561 = 9^4
So
9^24/9^4 Reminder is 0
Ans : 0 - 8 years agoHelpfull: Yes(27) No(1)
- sum of no. is 90 so its divisible by 3 0r 9. Similarly 6561 is also divisible by 3 or 9. S Remainder is 0.
- 8 years agoHelpfull: Yes(10) No(4)
- total sum 90 so (9+0=9) its factor is 3 or 9
then 6561=9^4
therefore 9^24/9^4 and their remainder is 0
so answer is 0 - 7 years agoHelpfull: Yes(2) No(0)
- Sorry I did mistake.
Ans is 2*2*2*2*2*1*1*1*1 =32 - 8 years agoHelpfull: Yes(1) No(10)
- 0
9^4=6561
then (9^4)^6 - 7 years agoHelpfull: Yes(0) No(0)
- {(9*x)^n}/(9^m) always leaves the reminder 0 if n>m and if n
- 5 years agoHelpfull: Yes(0) No(0)
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