Page 44 - ICSE Math 8

Example 5: Write the units digit of the squares of: (a) 61 (b) 372 (c) 2,134

Solution: (a) Since 61 has 1 at the units place, its square will have 1 as the units digit.
(b) Since 372 has 2 at the units place, its square will have 4 as the units digit
(c) Since 2,134 has 4 at the units place, its square will have 6 as the units digit.
Example 6: Write the Pythagorean triplet, whose members are formed by (a) 10 and (b) 16.

Solution: (a) m = 10 (b) m = 16
2m = 2 × 10 = 20 2m = 2 × 16 = 32
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2
2
2
m – 1 = (10) – 1 = 100 – 1 = 99 m – 1 = (16) – 1 = 256 – 1 = 255
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2
2
2
m + 1 = (10) + 1 = 100 + 1 = 101 m + 1 = (16) + 1 = 256 + 1 = 257
So, the Pythagorean triplet is So, the Pythagorean triplet is
20, 99, 101. 32, 255, 257.
Example 7: Without adding, find the following sum.
(a) 1 + 3 + 5 + 7 (b) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
Solution: (a) 1 + 3 + 5 + 7 is the sum of the first 4-odd natural numbers.
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So, 1 + 3 + 5 + 7 = (4) = 16.
(b) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 is the sum of the first 9-odd natural numbers.
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So, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = (9) = 81.
Example 8: Write (a) 25 and (b) 144 as a sum of odd natural numbers.
Solution: (a) 25 = 5 2
So, 25 can be written as a sum of the first 5-odd natural numbers.
\ 25 = 1 + 3 + 5 + 7 + 9
(b) 144 = (12) 2
So, 144 can be written as a sum of the first 12-odd natural numbers.
\ 144 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

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Example 9: Write 15 as a sum of two consecutive natural numbers.
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2
2
Solution: (2n + 1) = (2n + 2n) + (2n + 2n + 1) where 2n + 1 is an odd natural number.
15 = 2 × 7 + 1 {Compare 2 × 7 + 1 with 2n + 1}
Here n = 7
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2
2
\ 15 = (2 × 7 + 2 × 7) + (2 × 7 + 2 × 7 + 1)
= (2 × 49 + 2 × 7) + (2 × 49 + 2 × 7 + 1)
= (98 + 14) + (98 + 14 + 1) = 112 + 113
Example 10: Observe the following pattern:
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2
2
11 × (sum of digits of 11 ) = 11 × (1 + 2 + 1) = 22 2
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2
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111 × (sum of digits of 111 ) = 111 × (1 + 2 + 3 + 2 + 1) = 333 2
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2
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1111 × (sum of digits of 1111 ) = 1111 × (1 + 2 + 3 + 4 + 3 + 2 + 1) = 4444 2
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2
Now find the value of 111111 × (sum of digits of 111111 ).
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Solution: (111111) = 12345654321
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Sum of digits of (111111) = 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1
2
2
111111 × (sum of digits of 111111 )
2
= 111111 × (1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1) = 666666 2
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