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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows N is odd; so it's unit digit is odd. Therefore, the unit digit must be 1, 3, 5, 7 or 9.

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P Pankaj Sanodiya
The detailed solution for the above-written problem is as follows, The unit digit, when divided by 5, must be leaving a remainder of 3. So the unit digit must be either 3 or 8.

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A Amit Singh
If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11. let the number be 1089 here a = 1, b = 0, c = 8 and d = 9 1089 = 1000*1 + 100*0 + 10*8 + 9          = (1001*1 + 99*0 + 11*8) + [(0 + 9) - (1 + 8)]          = 11(91*1 + 9*0 + 8) + [ 9 - 9 ] here  [ (b + d) - (a + c) ] = [9 - 9 ] = 0 which is divisible by 11. hence If the number abcd is divisible by 11...

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P Pankaj Sanodiya
let the number abc be 132 Here a = 1, b = 3 and c = 2 132= 100*1 + 10*3 + 2 = 99 + 11*3 + (1 - 3 + 2)     = 11(9*1+3) + (1 - 3 + 2 )  if number is divisible by 11 then (a - b + c )  must be divisible by 11. as in above case of number 132 the a - b + c = 1 -3 + 2 = 0 whichis divisible by 11. Hence we conclude ( a - b + c ) should be divisible by 11 if abc is divisible by 11.  

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P Pankaj Sanodiya
Yes, it has been prooved that if a number is divisible by any number m, then it will also be divisible by each of the factors of m. Let's Assume n is divisible by m, and m is divisible by k.This means n=pm and m = qk where all are integers Now, n = p(qk) =( pq)k which means n is divisible by k. Hence a number is divisible by any number m, then it will also be divisible by each of the factors of m.

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P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 927 = 9 + 2 + 7 = 18 which is divisible by 3.  Hence we conclude number 108 is divisible by 3.

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P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 432 = 4 + 3 + 2 = 9 which is divisible by 3.  Hence we conclude number 432 is divisible by 3.

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P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 294= 2 + 9 + 4 = 15 which is divisible by 3.  Hence we conclude number 294 is divisible by 3.

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P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 616 = 6 + 1 + 6 = 13 which is not divisible by 3.  Hence we conclude number 616 is not divisible by 3.

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P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 108 = 1 + 0 +8 = 9 which is divisible by 3.  Hence we conclude number 108 is divisible by 3.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 927 = 9 + 2 + 7 = 18 which is divisible by 18. Hence we conclude number 927 is divisible by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digit 432 = 4 + 3 + 2 = 9 which is divisible by 9. Hence we conclude 432 is divisible by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 294 = 2 + 9 + 4 = 15 which is not divisible by 9. Hence we conclude 294 is not divisible by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 616 = 6 + 1 + 6 = 13 which is not divisible by 9, Hence we conclude 616 is not divisible by by 9.

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P Pankaj Sanodiya
Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digit of 108 = 1 + 0 + 8 = 9 which is divisible by 9 i.e (9/9 = 1). hence we conclude 108 is divisible by 9.

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P Pankaj Sanodiya
Since N leaves the remainder of 4 when divided by 5. the possible values in ones place of number N are 4 or 9. now, since it leaves a remainder of 1 when divided by 2, the N would be an odd number. hence ones digit of N is also an odd number. which means ones digit of our number N is 9.  

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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows N is Even; so it's unit digit is even. Therefore, the unit digit must be 2, 4, 6, 8 or 0.

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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows If the unit digit of a number is 0 or 5, then it is divisible by 5. hence if we need the remainder of 4 then unit digit of number should be 4 or 9.

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P Pankaj Sanodiya
The detailed solution for the above-written question is as follows If a number is divisible by 5 then it's unit digit must be 0 or 5. so if we need the remainder of 1 when divided by 5 then the numbers unit digit must be 1 or 6.

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P Pankaj Sanodiya
Let choosen number be abc then, abc = 100a + 10b + c cab = 100c + 10a + b bca = 100b + 10c + a The addition of all above three, abc + cab + bca = 111(a + b + c) = 37 × 3(a + b + c), which is divisible by 37 here a = 9, b = 3, and c = 7 937 + 793 + 397 = 2109 = 37*57  i.e. divisible by 37.
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