Question

There are N numbers of gold biscuits in the house, in which four people are lived. If the first men woke up and divided the biscuits into 5 equal piles and found one extra biscuit. He took one of those piles along with the extra biscuit and hid them. He then gathered the 4 remaining piles into a big pile, woke up the second person and went to sleep. Each of the other 3 persons did the same one by one, i.e., divided the big pile into 5 equal piles and found one extra biscuit. Each hid one of the piles along with the extra biscuit and gathered the remaining 4 piles into a big pile. If N $>$ 1000, what could be the least value of N?

• A. 1200
• B. 1346
• C. 1046
• D. 1246
• E. none

Answer 1

The correct option is D

1246

Suppose N=5x+1 A took (x+1) biscuit. Now 4x is of the form 5y+1 then x must be in the form 5z+4 $\Rightarrow$ 4(5z+4)=5y+1 $\Rightarrow$ y=4z+3 and x=5z+4 The ratio of number of biscuits that A and B took is [(5z+4)+1]:[(4z+3)+1]=5:4 So, we can say that any two successive persons A, B, C and D take coins in the ratio of 5:4 Let the number of biscuits that A,B,C and D took be a, b, c and d respectively. a:b=b:c=c:d=5:4 a:b:c:d=125:100:80:64 $\Rightarrow$ a=125k $\Rightarrow$x=125k-1 and N=5x+1=625k-4 As,N$>$1000, the least value of N is when k=2 $\Rightarrow$ N= 1246