Recently have seen quite a few questions on SUCCESSIVE REMAINDERS and not many people comfortable with them. i don't know how others solve it but the method I'll discuss here is the easiest one I guess. Suppose it has asked you to get the no which leaves remainder x,y,z when successively divided by a,b,c. All you need to do is write the divisor and respective remainder below it: a b c x y z
Now the no will be {[(b*z) + y] * a} + x
let's take an example, tell the no which leaves remainder 2,5,6 when successively divided by 6,8,11
6 8 11 2 5 6 no will be 8*6 = 48 +5 = 53 *6 = 318 +2 = 320
Hope it is clear, going to questions. 1) This was posted today in morning:
Find the sum of the remainders obtained when a number n is divided by 9 and 7 successively, if n is the smallest number that leaves respective remainders of 4, 6 and 9 when divided successively by 13, 11 and 15.
13 11 15 4 ...6...9
No will be= 11*9 = 99+6 = 105*13 = 1365 + 4 =1369 Rest part can be done easily.
2) K is a number which when successively divided 3 times by N leaves a remainder of 4,4,1.if NTo get the no's take N = 8,7,6,5 When 8 8 8 4 4 1 K = 8*1 = 8 +4 =12*8 = 96+4 = 100 similarly for 7,6,5. You can try for these values, and you will get K as 81, 64 , 49.
If there is something extra you want to add, please go ahead and do let me know if there is any flaw.
Two vessels A and B of equal capacities contain mixtures of milk and water in the ratio of 4:1 and 3:1 respectively. 25% of the mixture A taken out and added to B. After mixing it thoroughly,an equal amount is taken out from B and added back to A. The ratio of milk to water in vessel A after second operation is:
A man invested Rs. 1000 at the rate of 15% per annum at S.I. He withdraws the final amt. after 'T' years. He keeps half of the withdrawn amt with him and invests the remaining. This invested amt kept on reducing at S.I rate of 12.5% per annum for a period of 6 years. If the aggregate sum with the man after "T + 6" years is Rs. 1000, then find the value of 'T'
88888888888888888888......1000 8s divisible by 9,99,999,1,11,111,7,27 37 u get a remainder(these fundas are known) .similarly do u have a funda for 19 or any other numbers???
A circle is drawn with O as the center, Points A, B, C, D are chosen on the circle such that B and C lie on the same side of diameter AD. BC and AD when extended meet at point E and length of CE is equal to the radius of the circle. If length of arc AB is equal to length of arc BC, what is the measure of angle AEB to the nearest integer.
Options nai hai.. evn solution nai hai.. isko kaise karenge pls batao !!!!
In Δ ABD, ∠B = 60°, ∠D = 90° and AB = 4 cm. C is a point on BD, such that DC = DA. M is a point on CA (between C and A) such that 2CM = MA. N is the midpoint of BM. If Line CN intersects AB at P, find BP.
Consider the set G of all integers between 100 and 9999 (including the extremes). Call two integers a and b in G to be in the same category if the digits appearing in a and b are the same. For example, if a = 101, b = 100, c = 1000 and d = 120, then a, b and c are in the same category but c and d are not. Find the number of different categories that can be formed out of G.
pq, pr, qp and pp are two-digit numbers, 'pqr' is a three-digit number and 'prrp' is a four-digit number (p, q, r are all numerals). If pq × pr = pqr, then the remainder when (qp × prrp) is divided by 'pp' is
