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There are two Arithmetic Progressions A and B such that their nth terms are given by An = 101 + 3(n 1) and Bn = 150 + (n 1), where n is the set of natural numbers. The first 50 terms of A and B are written alternately i.e. A1B1A2B2..A50B50. What is the remainder when the number so formed is divided by 11?...

what can be the coolest approach to solve such problems...:-P

In such problems drz alwys sm kinda pattern.

here all d terms are of 3 digits and acc to div rule of 11 jst subtract the triplets adjacent to each odr and d diff will gv d remainder.

In dis q a1=101, b1=150, b1-a1=49, rem=5

a2=104, b2=151, b2-a2=47, rem=3

a3=107, b3=152, b3-a3=45, rem=1

a4=110, b4=153, b4-a4=43, rem= -1

continue in dis manr nd u will find d 1st 11 rem will sum out to b 0.

so for num till a44b44, rem wil b 0

nw we need to find the rem for rest of terms from a45,

a45=233, b45=194, diff=39, rem= -5

Here no need to solve further as u the pattern will b familier, the pattern of rem wil b lyk dis

-5, -3, -1, 1, 3 ,5

only 6 terms were left frm a45 to a50, so the will give the above rem.. and hence the total rem wil b zero.

dis approach is more of reasoning, so u need to give it a try by urself 4 its cmplt understanding...

Mine is coming 43.

Have u taken care that the pairs give birth in the 2nd month.

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Mine is coming 43.

Have u taken care that the pairs give birth in the 2nd month.

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1. A research conducted by BMC showed that each pair of rats gives birth to one pair of rats every month, starting from the 2nd month since it is born (e.g. A pair of rats born in March will give birth to one pair in May,June,July,etc).If there is only one pair of rats initially how many pairs of rats will there be at the end of 10 months?

66 pairs ? is it ? correct me pls

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