Question

Da 13 1. Find the greatest number that will exactly divide 840 and 2296 2. Find the greatest number that will divide each of 90. 108 and 125 without ang remainder 3. Find the smallest number that can be exactly divided by 72 and 108. 4. Find the smallest number of five ogts that can be eactly divided by 60.90 and 80 5. Find the greatest number of three ogts that can be easily divided by 75.45 and 60. 6. Find the greatest number that will divide 517 and 815 leaving the remainders 1 and respectively 7. Find the greatest number that will divide 37.56 and 93 leaving the remainders 1.2 and respectively 8. Find the greatest number that will divide 137, 182 and 422 leaving the remainder 2 in ea case 9. Find the smallest number which when divided by 6. 8. 12. 15 and 20 leaves the same remainder 5. 10. Find the greatest number of sex digits which is exactly divisible by 27.45.60.72 and 96 1. There is a heap of blocks. Anil makes separate heaps of 28 blocks from them. Shilpa makes separate heaps of 32 blocks from the same heap. Nitin makes separate heaps of 42 blods. Each of them finds that 5 blocks are left after making the separate heaps What is the smales​

Answer 1

Answer:

1. 8
3. Therefore, the smallest 5 digit number that is exactly divisible by 72 and 108 is 10152.
4. Hence, 10080 is the smallest five digit number which is exactly divisible by 60,80,90
5. 900
6. When we divide 517 and 815 by x the remainder is 1 and 3. Now, when we divide 516 and 812 by x the remainder will be 0. Therefore, x is the HCF of 516 and 812. Therefore, the greatest number will divide 517 and 815 leaving reminders 1 and 3 respectively is 4.
7. difference between (37-1)&(56-2)=18 which is factor of (93-3)=18
8. 15
9. 120 is the smallest number which will completely divide the numbers (6, 8, 12, 15, 20). Since, we want remainder 5.

10. Given numbers are 27 , 45 , 60 , 72 and 96

We have to find greatest 6-digit number which is divisible by given numbers.We first find lowest Common Multiple of given numbers then we find multiple of LCM which greatest 6 digit number.27 = 3 × 3 × 345 = 5 × 3 × 360 = 2 × 3 × 2 × 572 = 2 × 2 × 2 × 3 × 396 = 2 × 2 × 2 × 2 × 2 × 3LCM = 3 × 3 × 2 × 2 × 2 × 2 × 2 × 3 × 5 = 4320Now we find the multiples of 4320 which is largest 6 digit number for divide 999999 ÷ 4320 we get 231.48125So, 231st multiple of 4320 is largest 6 digit number .i.e., 4320 × 231 = 997920

• Therefore, required number is 997920

1. ​

28p + 5  = 32q + 5 = 42r +  5  = Total blocks

=> 28p = 32q = 42r  = Total blocks - 5

Lets find out LCM of 28 , 32 & 42

28 = 2 * 2 * 7

32 = 2 * 2 * 2 * 2 * 2

42 = 2 * 3 * 7

LCM = 2 * 2 * 2 * 2 * 2 * 3 * 7 =  672

672 =  total Blocks - 5

=> Total Blocks  = 677

smallest possible number of blocks in the heap = 677