【正文】 ger power can have only 2 last digits: 4,when the power is odd or 6 when the power is even. Hence, to get the remainder of 4^x/10 we should know whether the power is odd or even: if it39。s odd the remainder will be 4 and if it39。s even the remainder will be 6.(1) a = 1 4^(2a+1+b) = 4^(3+b) depending on b the power can be even or odd. Not sufficient.(2) b = 2 4^(2a+1+b) = 4^(2a+3) = 4^(even+odd)= 4^odd the remainder upon division of 4^odd by 10 is 4. Sufficient.Answer:B.3. Collection of GMATRemainder Problems這些題目妹紙覺得還是自己做一遍會很有用哦?。?！所以妹紙把分析全部隱藏在黃色highlight里面啦。大家加油~~~~ r is the remainder when the positive integer n is divided by 7, what is the value of r(1)When n is divided by 21, the remainder is an odd number(2)When n is divided by 28, the remainder is 3Sol：The possible reminders can be 1,2,3,4,5 and 6. We have the pinpoint the exact remainder from this 6 numbers.St 1: when n is divided by 21 ( 7 and 3) the remainder is an odd number.But it cannot be 7, 3 or 9 . Hence the possibilities are: 1 and 5.Hence there can be two remainders ,1 and 5, when divided by 7.NOT SUFFICIENTSt 2: when n is divided by 28 the remainder is 3.As 7 is a factor of 28, the remainder when divided by 7 will be 3SUFFICIENTAnswer: B2. If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?(1) n = 2(2) m = 1Sol：The Concept tested here is cycles of powers of 3.The cycles of powers of 3 are : 3,9,7,1St I) n = 2. This makes 3^(4*2 +2 + m) = 3^(10+m). we do not know m and hence cannot figure out the unit digit.St II) m=1 . This makes 3^(4*n +2 + 1).4n can be 4,8,12,16...3^(4*n +2 + 1) will be 3^7,3^11, 3^15,3^19 ..... in each case the unit digit will be 7. SUFFHence B p is a positive odd integer, what is the remainder when p is divided by 4 ?(1) When p is divided by 8, the remainder is 5.(2) p is the sum of the squares of two positive integers.Sol：st1. take multiples of 8....divide them by 4...remainder=1 in each case...st2. p is odd, since p is square of 2 integers...one will be even and other odd....now when we divide any even square by 4 vll gt 0 remainder..and when divide odd square vll get 1 as remainder......so in total remainder=1Ans : D p and n are positive integers and p n, what is the remainder when p^2 n^2 is divided by 15 ?(1) The remainder when p + n is divided by 5 is 1.(2) The remainder when p n is divided by 3 is 1.Sol：st1) p+n=6,11,16....insuff.st2) pn=4,7,10....insuff...multiply these two to get p^2n^2.....multiplying any two values from the above results in different remainder......also can be done thru equation....p+n=5a+1..and so onAns: E is the remainder when the positive integer x is divided by 3 ?(1) When x is divided by 6, the remainder is 2.(2) When x is divided by 15, the remainder is 2.Sol：st 1：multiple of 6 will also be multiple of 3 so remainder will be same as 2.st 2：multiple of 15 will also be multiple of 3....so the gives remainder 2 when divided by 15 also gives 2 as the remainder when divided by 3...Answer D is the remainder when the positive integer n is divided by 6 ?(1) n is a multiple of 5.(2) n is a multiple of 12.Sol：st 1) multiples of 5=5,10,15....all gives differnt remainders with 6st 2) n is divided by 12...so it will be divided by 6...remainder=0Answer B7.If x, y, and z are positive integers, what is the remainder when 100x + 10y + z is divided by 7?(1) y = 6(2) z = 3Sol：We need to know all the variables. We cannot get that from both the statements. Hence the answer is E.8.If n is a positive integer and r is the remainder when 4 +7n is divided by 3, what is the value of r ?(1) n + 1 is divisible by 3.(2) n 20Sol：st1) n+1 divisible by 3..so n=2,5,8,11......this gives 4+7n=18,39,60....remainder 0 in each case......st2) insufficient ....n can have any valueAnswer A9.If n is a positive integer and r is the remainder when (n 1)(n + 1) is divided by 24, what is the value of r ?(1) n is not divisible by 2.(2) n is not divisible by 3.Sol：ST 1 if n is not divisible by 2, then n is odd, so both(n 1) and (n + 1) are even. Moreover, since every other even number is a multiple of 4, one of those two factors is a multiple of 4. So the product (n 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x 2 x 2 = three 239。s in its prime factorization.But this is not sufficient, because it can be (n1)*(n+1)can be 2*4 where remainder is 8. it can be 4*6 where the remainder is 0.ST 2 if n is not divisible by 3, then exactly one of (n 1) and (n + 1) is divisible by 3, because every third integer is divisible by 3. Therefore, the product (n 1)(n + 1) contains a 3 in its prime factorization.Just like st 1 this is not sufficientST1+ST2 the overall prime factorization of (n 1)(n +1) contains three 239。s and a 3.Therefore, it is a multiple of 24.SufficientAnswer