Type # 1.

find smallest no. other than k, that leaves remainder k when divided by w,x,y...

to solve such questions, take lcm of w,x,y...and add k to it.

e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...

take lcm of 6,7,8,9 and add 4

i.e. 504 + 4 = 508

Type # 2

find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.

unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1

in such questions, take lcm of divisors n subtract the common difference from it

here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23

Type # 3

Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,

we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208

now we have one more condition...remainder 1 with 11.

concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.

i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.

so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1

208%11 = 10

210k%11 = k

therefore, 10 + k shud leave remainder 1 when divided by 11.

hence, k = 2. and the no. is 208 + 210 x 2 = 628

e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7

for first 3 conditions....no. is 120 + 2 = 122

hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362

Type # 4

What if there is no relation between divisors n remainders?

e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.

we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant.

in such cases...take 1 case n target another case...

e.g. i take the case 7 with 13...and target 6 with 11.

which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.

now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6.

a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72

now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.

to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...

hence the no. is of the form 72 + 143 k.

72 + 143k % 7 = 2 + 3k

now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..

a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.

hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.

now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.

For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically.

there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same...

ORIGINALLY POSTED BY MAXIMUS

RC practice :- http://codecoax.com/grerc/