The smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case

Prime factors of 4, 7 and 13

4 = 2x2

7 and 13 are prime numbers.

LCM ( 4, 7, 13) = 364

We know that, the largest 4 digit number is 9999

Step 1: Divide 9999 by 364, we get

9999/364 = 171

Step 2: Subtract 171 from 9999

9999 - 171 = 9828

Step 3: Add 3 to 9828

9828 + 3 = 9831

Therefore 9831 is the number.

The smallest number which when divided by 20,25,35 and 40 leaves a remainder of 14,19,29 and 34 respectively, is 1394 . How?

Solution

In this type of questions , we first find LCM of given numbers and then subtract common value to get the answer.

Common value = Number - Remainder
So,
20–14 =6
25–19=6
35–29=6
40–34 = 6

Now LCM of 20, 25, 35 and 40 -

20 = 2^2 * 5

25 = 5^2

35 = 5*7

40 = 2^3 * 5

therefore LCM = 2^3 * 5^2 * 7 = 1400

As i wrote above to get the final answer we have to subtract common value from LCM.

Final answer = 1400 - 6 = 1394

Solution

Solve for LCM of given numbers

The prime factors of 17=17×1

The prime factors of 23=23 ×1

The prime factors of 29=29×1

∴HCF(17 ,23,29)=1

LCM(17,23,29)=17×23×29

∴LCM(17,23,29)=11339

Hence, the L.C.M of 17,23 and 29 is 11339 and H.C.F. of 17,23 and 29 is 1.