Olympiad Test: Playing With Numbers - 2 - Class 8 MCQ

For a number to be divisible by 9, its digits sum should be divisible by 9.
∴ 7 + 6 + 3+*+ 3+1+2 = 22 +*.
∴ 5 should be written on place of * to make the number divisible by 9. 

⇒ Sum of digits at odd places
= * + 1 + 6 = * +7
Sum of digits at even places
= 5 + 2 + 7
= 14 
For divisibility with 11, *+ 7 = 14
⇒ * = 7

3c = x B, where, x is the carry,
∴ x + 3B = y B, where y is the carry.
∴ y + 3A = B.
∴ C = 8, B = 4, A = 1.

Sum of first 22 even natural numbers
= (2 + 4 + …… + 44) 
= 2 (1 + 2 + …… + 22)

= 2 × 11 × 23
= 22 × 23
= 506

Let the candles have burnt for ‘x’ hours,

⇒ 

735 is a number having all its digits, a prime number and all the digits of 735 are the factors of 735.

We have to find, the value of
(a+b+c+d +e)+(x+ y + z)+ f = x
∴ Sum of numbers of any row/column = 34
∴ a +b+13+6= x+ y + z +13
= e+b +7 +9=c + d +6+ f
= 5 + 16 + a + 2 = x +e+d + 5
= 16 + y + 7 +c= 2 + z + 9 + f = 34
⇒ 2(a+b+c+d +e+ f + x+ y+ z)
+19 + 13 + 16 + 6 + 23+ 5
+23 + 11 34 x 8
⇒ 2x + 116 = 272
⇒ x = 78

Let the natural numbers be, y and z.

⇒
⇒ 
⇒ xyz ≤ 37.03
∴ The limiting value of xyz is 37.03,
∴ The numbers x,y and z are natural.
∴ 37.03 cannot be obtained as a product.
∴ 37 is a prime number.
∴ 36 is the greatest number which can be obtained as a product of 3 natural numbers whose sum is 10. 

60 = 1
6¹ = 6
6² = 36
6³ = 216
6⁴ = 1296 
6⁵ = 7776
∴ It is observed that 6ⁿ has 6 at its units place.
∴ 6²²² has 6 as its unit’s place  digit.

(a) 1729 = (12)³ + (1)³= (10)³ + (1)³
∴ 1729 can be expressed as sum of two perfect natural cubes.
It is the smallest number to satisfy this condition, and, is known as Ramanujan’s Number.

x,y are respective carries.
y + P + E + F E
x + 2A = y E
T + T = x A
⇒ P = 9, A = 8, T = 4, E = 6 and F = 1
∴ F = 1

 xyz = x + y + z …(i) We also know that,

∴ Sum of 3 natural numbers > 3.
∴ From general interpretation,
1 + 2  + 3 = 1 ⇒ 2 ⇒ 3
∴ Required set of natural numbers = (1, 2, 3).

∵ Last (unit) place digit = 5
∴ The least natural number whose perfect square is having 5, as its unit place digit will be equal to 5.
∴ 52 = 25

135 × 135
Trick :
135 × 135 = 25
Multiply the units place 5 and write the product, Add 1 to any one of the remaining digits at tenths place and write the product on tenths place.
13 × 14 = 182
∴ 135 × 135 = 18225

∵ Perfect squares should have 1, 4, 9, 6 and 5 as their units place digit.
∴ 1282 is not a perfect square. 

Interchange of ×, ÷ will produce the desired result.

Least number which is divisible by 2, 3, 5 and 55
= LCM 2, 3, 5 and 55
= 66 × 5
= 330

60 < Sum of two triangular numbers = perfect square. The smallest number satisfying this condition is 64.

9¹ = 9
9² = 81
9³ = 243 × 3 = 729 
9⁴ = 6561
9⁵ = 59049
∴ 9⁴ⁿ = 6561
∴ 9²⁰⁰ has 1 as its units place digit.
∴ 9²⁰¹ has 9 as its units place digit. 

P = 8, Q = 7, R = 6 
∴ P = 8