pairs of 2 and 5. Therefore, 90 should be multiplied by 2 x 5, i.e., 10.
Hence, the required square number is 90 x 10=900.
EXERCISE 6.3
1. What could be the possible 'one's' digits of the square root of each of the following
numbers?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
2. Without doing any calculation find th
not nerfect squares
Answers
Answer:
Step-by-step explanation:
Solution: We have, 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7
If we divide 9408 by the factor 3, then
9408 ÷ 3 = 3136 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 which is a perfect square. (Why?)
Therefore, the required smallest number is 3.
And, 3136 = 2 × 2 × 2 × 7 = 56.
2 6, 9, 15 Example 8: Find the smallest square number which is divisible by each of the numbers
3 3, 9, 15 6, 9 and 15.
3 1, 3, 5 Solution: This has to be done in two steps. First find the smallest common multiple and
5 1, 1, 5 then find the square number needed. The least number divisible by each one of 6, 9 and
1, 1, 1 15 is their LCM. The LCM of 6, 9 and 15 is 2 × 3 × 3 × 5 = 90.
Prime factorisation of 90 is 90 = 2 × 3 × 3 × 5.
We see that prime factors 2 and 5 are not in pairs. Therefore 90 is not a perfect
square.
In order to get a perfect square, each factor of 90 must be paired. So we need to
make pairs of 2 and 5. Therefore, 90 should be multiplied by 2 × 5, i.e., 10.
Hence, the required square number is 90 × 10 = 900.
Answer:
I dodny understand the fact