If N<1000, how many biscuits were left after the fourth man took his share?
624
621
252
257
Explanation:
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
⇒4(5z+4)=5y+1
⇒y=4z+3andx=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
⇒a=125k
⇒x=125k−1 and N=5x+1=625k−4
⇒N<100, then k=1
⇒N=621
⇒621=(5×124)+3
4×124=(5×99)+1
4×99=(5×79)+1
4×79=(5×63)+1
After the fourth man took his share (5×63+1), the biscuits lefts is 4×63= 252
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
⇒4(5z+4)=5y+1
⇒y=4z+3andx=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
⇒a=125k
⇒x=125k−1 and N=5x+1=625k−4
⇒N<100, then k=1
⇒N=621
⇒621=(5×124)+3
4×124=(5×99)+1
4×99=(5×79)+1
4×79=(5×63)+1
After the fourth man took his share (5×63+1), the biscuits lefts is 4×63= 252