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(1Ã—2)+(2Ã—3)=

(1Ã—2)+(2Ã—3)+(3Ã—4)=

(1Ã—2)+(2Ã—3)+(3Ã—4)+(4Ã—5)=

(1 Ã— 2) + (2 Ã— 3) + (3 Ã— 4) + (4 Ã— 5) + (5 Ã— 6)

Answer 8: T

If the biggest number (factor) on the LHS is 3, the multiplication of the three numbers on the RHS begins with 2.

If the biggest number (factor) on the LHS is 4, the multiplication of the three numbers on the RHS begins with 3.

If the biggest number (factor) on the LHS is 5, the multiplication of the three numbers on the RHS begins with 4.

Using this pattern, (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5) + (5 x 6) has 6 as the biggest number. Hence, the multiplication of the three numbers on the RHS will begin with 5. Finally, we have:

1 Ã— 2 + 2 Ã—3 + 3 Ã— 4 + 4 Ã— 5 + 5 Ã— 6 = = 70

1=1/2 {1Ã—(1+1)}

(i) 1 + 2 + 3 + 4 + 5 + ... + 50

(ii) 31 + 32 + ... + 50

The first equality, whose biggest number on the LHS is 1, has 1, 1 and 1 as the three numbers.

The second equality, whose biggest number on the LHS is 2, has 2, 2 and 1 as the three numbers.

The third equality, whose biggest number on the LHS is 3, has 3, 3 and 1 as the three numbers.

The fourth equality, whose biggest number on the LHS is 4, has 4, 4 and 1 as the three numbers.

Hence, if the biggest number on the LHS is

Using this property, we can calculate the sums for (i) and (ii) as follows:

(ii) The sum can be expressed as the difference of the two sums as follows:

The result of the first bracket is exactly the same as in part (i).

Then, the second bracket:

Finally, we have:

(i) 1

(ii) 5

The first equality, whose biggest number on the LHS is 1, has 1, 1, 1, 2, 1 and 1 as the six numbers.

The second equality, whose biggest number on the LHS is 2, has 2, 2, 1, 2, 2 and 1 as the six numbers.

The third equality, whose biggest number on the LHS is 3, has 3, 3, 1, 2, 3 and 1 as the six numbers.

The fourth equality, whose biggest number on the LHS is 4, has numbers 4, 4, 1, 2, 4 and 1 as the six numbers.

Note that the fourth number on the RHS is always 2 and the sixth number is always 1. The remaining numbers are equal to the biggest number on the LHS.

Hence, if the biggest number on the LHS is n, the six numbers on the RHS would be n, n, 1, 2, n and 1.

Using this property, we can calculate the sums for (i) and (ii) as follows:

(i) 1

1

= 1/6 Ã—[4Ã—(4+1)Ã—(2Ã—4+1)]

=1/6 Ã— 4 Ã— 5 Ã— 9 = 30

Finally, the wanted sum is: 52 + 62 + ...... + 122= (12 + 22 +..... + 122) âˆ’ ( 12 + 22 +..... + 122)=650 âˆ’ 30 = 62

121, 225, 256, 324, 1296, 6561, 5476, 4489, 373758

(i) 256

= (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2)

There are no factors that are not paired. Hence, 256 is a perfect square. The square of an even number is always even. Hence, 256 is the square of an even number.

(ii) 324

= (2 x 2) x (3 x 3) x (3 x 3)

There are no factors that are not paired. Hence, 324 is a perfect square. The square of an even number is always even. Hence, 324 is the square of an even number.

(iii)1296

= (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3)

There are no factors that are not paired. Hence, 1296 is a perfect square. The square of an even number is always even. Hence, 1296 is the square of an even number.

(iv) 5476

= (2 x 2) x (37 x 37)

There are no factors that are not paired. Hence, 5476 is a perfect square. The square of an even number is always even. Hence, 5476 is the square of an even number.

(v) 373758

Here, each factor appears only once, so grouping them into pairs of equal factors is not possible. It means that 373758 is not the square of an even number.

Hence, the numbers that are the squares of even numbers are 256, 324, 1296 and 5476.

(i) 1026

(ii) 1028

(iii) 1024

(iv) 1022

(v) 1023

(vi) 1027

(i) 1026 has 6 as the units digit, so it is possibly a perfect square.

(ii) 1028 has 8 as the units digit, so it cannot be a perfect square.

(iii) 1024 has 4 as the units digit, so it is possibly a perfect square.

(iv) 1022 has 2 as the units digit, so it cannot be a perfect square.

(v) 1023 has 3 as the units digit, so it cannot be a perfect square.

(vi) 1027 has 7 as the unit digit, so it cannot be a perfect square.

Hence, by examining the units digits, we can be certain that 1028, 1022, 1023 and 1027 cannot be whole squares.

On the other hand, a number whose unit digit is 1, 4, 5, 6, 9 or 0 might be a perfect square (although we will have to verify whether it is a perfect square or not).

Applying the above two conditions, we cannot quickly decide whether the following numbers are squares of any numbers:

1111, 1444, 1555, 1666, 1999

On the other hand, a number whose unit digit is 1, 4, 5, 6, 9 or 0 might be a perfect square although we have to verify that.

Applying these two conditions, we cannot determine whether the following numbers are squares just by looking at their unit digits:

1111, 1001, 1555, 1666 and 1999

(i) The number of digits in a square number is even.

(ii) The square of a prime number is prime.

(iii) The sum of two square numbers is a square number.

(iv) The difference of two square numbers is a square number.

(v) The product of two square numbers is a square number.

(vi) No square number is negative.

(vii) There is no square number between 50 and 60.

(viii) There are fourteen square number upto 200.

Example: 100 is the square of a number but its number of digits is three, which is not an even number.

(ii) False

If

(iii) False

1 is the square of a number (1 = 1

(iv) False

4 and 1 are squares (4 = 2

(v) True

If

(vi) True

The square of a negative number will be positive because negative times negative is positive.

(vii) True

7

(viii) True

14