Page 26 - SM inner class 8.cdr

Step V: To ob tain the num ber whose square Example 4: 2500 is a perfect square as number
is the given num ber taken over one of zeros are 2(even) and 25000 is not a perfect
factor from each pair and multiply square as the number of zeros are 3 (odd).
them. Property 3: Squares of even numbers are

Example 1. Is 225 a perfect-square? If so, find always even numbers and square
the number whose square is 225. of odd numbers are always odd.
2
´
=
Solution: Resolving 225 into prime factors, we Example 5: 12 = 12 12 144 (both are even
2
´
obtain numbers) 19 = 19 19 = 361 (both are odd
225 = 3 ´ 3 ´ 5 ´ 5 numbers)
Grouping the factors in to pairs of like factors Property 4:
we have– Pythagorean Triplets: 2m, m - 1, m + are
2
2
1
´
5
225 = (3 ´ ) (5 ´ ) called pythagorean triplets, where 2 m is the
3
Clearly, 225 can be grouped into pairs of smallest even number. e.g. consider the
equal factors and no factor is left over. smallest number = 6 2, m = 6 so. m = 3
Hence, 225 is a perfect-square. Pythagorean Triplets are: 2m = 2 ´ 3 = 6
Again, 225 = (3 ´ ) (3 ´ ) m - = 3 2 - = 9 - = 8
2
5
´
5
1
1
1
´
=15 15 15 2 m + 1 = 3 3 + 1 = 9 + =10
=
2
1
or 225 = 3 ´ 3 ´ 5 ´ 5 So, 6, 8, 10 are pythagorean triplets.
{ {
If a, b and c are three natural numbers with
= 3 ´ 5 =15
‘a’ as the smallest of them, then
So, 225 is the square of 15.
(i) If a is odd, sum of other two numbers is
Example 2. Is 150 a perfect-square? If so,find equal to a e.g. 3, 4, 5; we have 3 = 4 + 5
2
2
the number whose square is 150.
(ii) If a is even, sum of other two numbers is
Solution: Resolving 150 into prime factors, we
8 2
obtain e.g. 8, 17, 15; we have = 17 + 15
2
150 = 2 ´ 3 ´ 5 ´ 5
2
2
Example 6: (i) For m=3, (2m, m - 1, m + )
1
Grouping the factors in to pairs of like factors
we have – will be (6,8,10 ) is a Pythagorean triplet.
2
2
(ii) For m=4, (2m, m -1, m + ) will be (8,15,
1
225 = 2 ´ 3 ´ ( 5 ´ 5 )
17 ) is a Pythagorean triplet.
Clearly, 150 can not be grouped into pairs of
equal factors.2 and 3 factors are left over. Example 7: If the smallest number of
Pythagorean triplet is 14 then find other two
Hence, 150 is not a perfect-square.
numbers.
Properties of Perfect Square Solution: We know that the Pythagorean
2
1
,
Property 1: A number having 2, 3, 7 or 8 at triplet is (2m m, 2 - 1 m + )
unit’s place is never a perfect So, 2m = 14
square. In other words, no square so m = = 7
number ends in 2, 3, 7 or 8.
Therefore, the other two numbers are
Example 3. None of the numbers 152, 7693, m - and m + 1 = 7 2 -1 and 7 + 1
2
2
2
1
14357, 88888, 798328 is a perfect square
because the unit digit of each number ends = 48 and 50
with 2,3,7 or 8. Prop erty 5: The square of a nat u ral num ber
Property 2: A number ending in an odd n is equal to the sum of the first
number of zeros is never a n odd num bers.
perfect square.
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Mathematics In Focus - 8