Page 144 - Ganit Kaushal

In diagram (ii), the number of pegs and the thread gap, that is 16 and 6, are not co-prime. Their HCF = 2, so
the thread ultimately passes after a gap of two pegs.

In diagram (iii), the number of pegs and the thread
gap, i.e., 12 and 4, are also not co-prime. Their Remember
HCF is 4. So, the thread passes after a gap of 4 1. 2 is the smallest prime number and the only one
pegs. (Note that 2 and 4 both are common divisors, which is even. Every prime number except 2 is odd.
but we take the greater one.) 2. 1 is co-prime with every number.

Therefore, we conclude that – if the number of 3. Any two successive numbers are always co-prime.
pegs and the thread-gap are co-prime, the 4. The sum of any two co-prime numbers is always
co-prime with their product.
thread ultimately passes through every peg. And 5. The two numbers having their units digits as 0 and
if they are not co-prime, the thread will pass 5 are never co-prime to each other.
through a gap of pegs that is equal to their HCF.

Solved Examples

Ex 1. In a jump jackpot game, what jump size can reach both 15 and 30? There are multiple jump sizes
possible; write them all. (NCERT)
Sol. 1, 3, 5, 15.  These are four common factors of 15 and 30.

Ex 2. In the treasure hunting game, Dumpy has kept treasures on 28 and 70. What jump sizes will land on
both the numbers? (NCERT)
Sol.  Factors of 28 = 1, 2, 4, 7, 14, 28 and Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70.
Common factors of 28 and 70 = 1, 2, 7, and 14.
\ Jump sizes that will land on the numbers 28 and 70 = 1, 2, 7, and 14.

Ex 3. Which of the following numbers are co-prime?
(a) 18 and 35 (b) 17 and 68
Sol. (a) Factors of 18 = 1, 2, 3, 6, 9, 18 and Factors of 35 = 1, 5, 7, 35.
We see that the HCF of 18 and 35 = 1. So, they are co-prime numbers.
(b) 17 and 68 are not co-prime.  HCF of 17 and 68 = 17.

Ex 4. A number is divisible by both 5 and 12. By which other number will that number always be divisible?
Sol. It will be divisible by the LCM of 5 and 12 i.e., 5 × 12 = 60 ( 5, 12 are co-prime).
\ The number will always be divisible by 60. It will be divisible by all the divisors of 60, too.
Ex 5. Find the smallest number that is a multiple of all the numbers from 1 to 10 2 1, 2, 3, 4, 5, 6, 8, 9, 10
except for 7. (NCERT) 2 1, 1, 3, 2, 5, 3, 4, 9, 5

Sol. We need to find the LCM of 1, 2, 3, 4, 5, 6, 8, 9 and 10. 2 1, 1, 3, 1, 5, 3, 2, 9, 5
Using the algorithm to find the LCM by common division, we obtain that: 3 1, 1, 3, 1, 5, 3, 1, 9, 5
3 1, 1, 1, 1, 5, 1, 1, 3, 5
LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360 5 1, 1, 1, 1, 5, 1, 1, 1, 5
\ 360 is the smallest number that is a multiple of all the numbers from 1, 1, 1, 1, 1, 1, 1, 1, 1
1 to 10 except for 7.
Ex 6. Find the smallest number that is a multiple of all the numbers from 1 to 10. (NCERT)
Sol. From Ex 5 above, we have the LCM of all the numbers from 1 to 10, except for 7 = 360.

\ LCM of all numbers from 1 to 10 = 360 × 7 = 2520. ( 7 is co-prime to all nos. from 1 to 10)

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