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(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Answer :

The squares of a number would be odd number depending upon the unit's digit of the number.If the number has an even unit digit then it will have an even square number and if the unit's digit is odd then it will have an odd square number.Here we have:

(i) \(431^2\) is an odd number as the unit's digit is odd.

(ii) \(2826^2\) is an even number as the unit's digit is even.

(iii) \(7779^2\) is an odd number as the unit's digit is odd.

(iv) \(82004^2\) is an even number as the unit's digit is even.

- 1.What will be the unit digit of the squares of the following numbers? (i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 20387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555
- 2. The following numbers are not perfect squares. Give reason. (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
- 4.Observe the following pattern and find the missing digits. \(11^2\) = 121 \(101^2\) = 10201 \(1001^2\) = 1002001 \(100001^2\) = 1…2…1 \(10000001^2\) = ………
- 5.Observe the following pattern and supply the missing numbers. \(11^2\) = 121 \(101^2\) = 10201 \(10101^2\) = 102030201 \(1010101^2\) = ………. \(……….^2\) = 10203040504030201
- 6.Using the given pattern, find the missing numbers. \(1^2 + 2^2 + 2^2 = 3^2\) \(2^2 + 3^2 + 6^2 = 7^2\) \(3^2 + 4^2 + 12^2 = 13^2\) \(4^2 + 5^2 + ….^2 = 21^2\) \(5^2 + ….^2 + 30^2 = 31^2\) \(6^2 + 7^2 + …..^2 = ……2\)
- 7.Without adding, find the sum. (i) 1 + 3 + 5 + 7 + 9 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
- 8. (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers.
- 9.How many numbers lie between squares of the following numbers? (i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100.

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