If 34a68 is to be divisible by 9, then the sum of all digits must be divisible by 9.

3 + 4 + a + 6 + 8 = multiple of 9

21 + a= multiple of 9 Thus, the possible value of a is 6.

5.

On Teacher's Day, sweets were to be equally distributed amongst 540 children. But on that particular day 135 children remained absent, hence each child got 2 sweets extra. How many sweets was each child originally supposed to get

The divisor is 25 times the quotie remainder. If the quotient is 16, thent and 5 times the dividend isamongst 540 children. But on that particular day 135 children remained absent, hence each child got 2 sweets extra. How many sweets was each child originally supposed to get?

We are not able to find any quicker method for this question. We suggest that you understand the working of its detailed solution.

Total zeroes at the end are produced due to three factors:

1)Due to zeroes at the end of numbers.

2)Due to 5 at the ten-place of numbers followed by zero at the unit place.

3)Due to 5 at the hundred-place of numbers followed by zeroes at the end.

Due to factor (1):

Number of numbers with single zero= 90 Number of numbers with double zeroes= 9 Number of numbers with triple zeroes= 1 Thus, the total no. of zeroes due to this factor

= 90 X 1 + 9 X 2 + 1X 3 = 111

Due to factor (2):

When 5 is multiplied by any even number, a zero is produced. There are 10 numbers which have 5 at the ten-place followed by zero (like 50, 150,250, ...., 950).

Thus, 10 zeroes are produced due to factor (2).

Due to factor (3): Following the same reasoning, we may say that 500 x 200 will produce an extra zero.

Thus, there are total Ill + 10 + 1 = 122 zeroes.

8.

In a question ; on ;division ;with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. Find the correct quotient.

For the number to be divisible by 11, difference of sums of digits at odd and even places should be multiple of 11. Here the difference is 40 - 26 = 14, which should be 11. As the last digit of N is at odd place, so if we reduce the digit at 21st place by 3 the number will be divisible by 11. In other words, the number N is 3 more than the multiple of 11. That is, if we divide the number by 11 we get a remainder 3.

10.

When a certain number is divided by 223, the remainder is 79.

When that number is divided by 179, the quotient is 315; then what will be the remainder?