Question

if a number is divided by x then the remainder is 8 and when 1/3 of that number is divided by x then remainder is 29. what is the minimum number greater than 1000.​

Answer 1

Answer:

Let the original number being divided be xx . Using the given statements we can write down the following equations :

x=kN+8x=kN+8

and, x3=mN+31x3=mN+31

⇒⇒ x=3mN+93x=3mN+93

Since, x=xx=x , we can further conclude that :

kN+8=3mN+93kN+8=3mN+93

⇒⇒ (k−3m)N=85(k−3m)N=85

Now, 85=5×1785=5×17

So, there are only 22 ways to write it as a product of two naturals, which are :;

1×851×85 and 5×175×17

But, we know that the number NN has to be greater than 3131 as it leaves a remainder 3131 in the second case. Thus, the only possibility is :

(k−3m)=1(k−3m)=1 and N=85N=85

Now, we can write x=85k+8x=85k+8

Given, x>1000x>1000

85k+8>100085k+8>1000

or, k>99285k>99285

or, k≥12k≥12

We also have to make sure that the kk is of the form 3m+13m+1 , as (k−3m)=1(k−3m)=1

So, we’ll take the value k=13k=13

Then, x=85×13+8x=85×13+8

⇒⇒ x=1105+8x=1105+8

⇒⇒ x=1113x=1113

Answer 2

Given :- if a number is divided by x then the remainder is 8 and when 1/3 of that number is divided by x then remainder is 29. what is the minimum number greater than 1000. ?

Answer :-

Let us assume that, the number is y .

so,

→ y ÷ x = 8 remainder .

then,

→ y = mx + 8 -------- (1)

again,

→ (y/3) ÷ x = 29 remainder .

→ (y/3) = nx + 29

→ y = 3nx + 87 -------- (2)

comparing both,

mx + 8 = 3nx + 87

→ mx - 3nx = 87 - 8

→ x(m - 3n) = 79

since 79 is a prime number

→ x(m - 3n) = 1 * 79 or 79 * 1

but since remainder in second case is 29 .

→ x > 29 => x = 79

then,

→ m - 3n = 1 .

therefore,

→ y = 79m + 8

so,

→ 79m + 8 > 1000

putting m = 13,

→ 1035 > 1000 .

Hence, 1035 is the required number .

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