model 2 divisibility, multiples, add and subtract based number system Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

Here, the first divisor (221) is a multiple of second divisor (13)

Hence, required remainder = remainder obtained on dividing 64 by 13 = 12

Answer: (d)

$(17)^200$ = $(18 –1)^200$

We know that

$ (x + a)^n$ =$x^n + nx^(n–1).a + {n(n - 1)}/{1 × 2}x^{n - 2}a^2 +{n(n - 1)(n - 2)}/{1 × 2 × 3}x^{n - 3}a^3+…+a^n$

We see that all the terms on the R.H.S. except $a^n$ has x as one of its factor and hence are divisible by x. So,$(x +a)^n$ is divisible by x or not will be decided by $a^n$ . Let x = 18, a = – 1 and n = 200

∴ $(18 –1)^200$ is divisible by 18 or not will depend on $(–1)^200$ as all other terms in its expansion will be divisible by 18 because each of them will have 18 as one of their factors.

$(–1)^200$ = 1 (Since 200 is even)

1 is not divisible by 18 and is also less than 18.

∴1 is the remainder

Answer: (c)

If the quotient in the first case be x.

Then, number = 5x + 3

On Squaring, the number

= $(5x + 3)^2$

= $25x^2$ + 30x + 9

On dividing by 5, remainder

= 9 – 5 = 4

Answer: (d)

Answer: (c)

The remainder will be same. On dividing 9 by 6, remainder = 3 On dividing 81 by 6, remainder = 3