model 2 divisibility, multiples, add and subtract based number system Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

$a^4 - b^4 = (a - b) (a + b) (a^2 + b^2)$,

Where a and b are odd positive integers.

If two positive integers are odd, then their sum, difference and sum of their squares are always even.

∴ (a - b) (a + b) and $(a^2 + b^2)$ are divisible by 2.

Hence (a - b) (a + b) x $(a^2 + b^2) = a^4 - b^4$ is always divisible by $2^3 = 8$

Answer: (c)

Answer: (b)

Here, the first divisor (91) is a multiple of the second divisor (13).

∴ Required remainder = Remainder obtained on dividing 17 by 13

⇒ 17 = ( 13 × 1 ) + 4

Hence Required remainder = 4

Answer: (b)

When the second divisor is a factor of the first divisor, the second remainder is obtained by dividing the first remainder by the second divisor.

Hence, on dividing 29 by 8, the remainder is 5.

Answer: (a)

In this problem, put any even positive value for both m & n,

For example m = 4 & n = 2

∴ (m² – n²) = (4² – 2²) = 16 - 4

= 12 is always divisible by 4.

Answer: (b)

First number (a) = 1125, Last number (an) = 4950,

Divisible by (d) = 225, No. of terms (n) = ?

No. of term formula, an= a+(n-1)d

Then, 4950= 1125 + (n - 1) 225

3825= 225n - 225

4050= 225n

n = 4050/225 = 18

Therefore, 18 numbers between 1000 and 5000 are exactly divisible by 225.