【正文】 ,22,25,28,31,34Common values are t = 1,13,25,2. Equate t = 4k+1 to t=3s+1We have 4k + 1 = 3s+1k = 3s/4since, k is an integer so only those values of s which are multiple of 4 will satisfy both STAT1 and STAT2so, mon values are given by t = 3s + 1 where s is multiple of 4so t = 1,13,25 (for s=0,4,8 respectively)Clearly we cannot find a unique reminder when t is divided by 5 as in some cases (t=1) we are getting the reminder as 1 and in some(t=10) we are getting the reminder as 0.So, INSUFFICIENTSo, answer will be EExample 2：If 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 1Sol:STAT1 : The remainder when p + n is divided by 5 is 1.p+n = 5k + 1but we cannot say anything about p^2 n^2 just from this information.So, INSUFFICIENTSTAT2 : The remainder when p n is divided by 3 is 1pn = 3s + 1but we cannot say anything about p^2 n^2 just from this information.So, INSUFFICIENTSTAT1+STAT2：p^2 n^2 = (p+n) * (pn) = (5k + 1) * (3s + 1)= 15ks + 5k + 3s + 1The reminder of the above expression by 15 is same as the reminder of 5k + 3s + 1 with 15 as 15ks will go with 15.But we cannot say anything about the reminder as its value will change with the values of k and s.So INSUFFICIENTHence answer will beDPractice:What is the remainder when positive integer t is divided by 5?(1) When t is divided by 4, the remainder is 1(2) When t is divided by 3, the remainder is 1B.方法二：重建法Given that an integer n when divided by an integer a gives r as reminder then the integern can be written asn = ak + rwhere k is a constant integer.Example 1： (C) 48 (趕時間的童鞋可以略過。Collection of MethodsMethod 1：小數(shù)法（妹紙自己取的名字，包括后面的方法也都是妹紙取的~歡迎討論）A way that the GMAT will test remainders is what you would typically just divide back into the problem to determine the decimals:25/4 = 6 remainder 1Divide that 1 back by 4 to get , so the answer is .Any number with a remainder could be expressed as a decimal.The remainder provides the data after the decimal point, and the quotient gives you the number to the left of the decimal point.Consider this problem (which appears courtesy of GMAC):Example： (B) 75 What is the remainder when B is divided by 6 if B is a positive integer?(1) When B is divided by 18, the remainder is 3(2) When B is divided by 12, the remainder is 9這題請大家自己試一試哦。EExample 3：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) n20.再次~~~請大家來考驗下自己吧~~~O(∩_∩)O~~Sol:STAT1: The remainder when x is divided by 7 is 2.x = 7k + 2Possible values of x are 2,9,16,...,51,...we cannot say anything about the values of xso, INSUFFICIENTSTAT2: The remainder when x is divided by 3 is 2.Most GMAT remainder problems are encountered inlimited toUsually you get 1, maximum 2 questions on remainders on thetest (based on GMAT Prep CATs)以下Pattern并不是按照重要性順序來排的喲~~~Pattern 6 is the most mon pattern! 妹紙覺得大家都可以看一看。(D) 20=9 y=75.Answer:Divisor x would be the least mon multiple of above two divisors 5 and 7, hence x=35.Remainder r would be the first mon integer in above two patterns, hence r=31.Therefore general formula based on both statements is n=35m+ the smallest positive integer k such that k+n is a multiple of 35 is 4 n+4 = 35k+31+4 = 35(k+1).Answer:Since x, the number of cards, should satisfy both conditions then it equals to 101.Answer:APattern5:Most mon patter.Q1：What is the remainder when the positive integer n is divided by 6?(1) n is multiple of 5(2) n is a multiple of 12Sol：(1) n is multiple of 5. If n=5, then n yields the remainder of 5 when divided by 6 but if n=10, then n yields the remainder of 4 when divided by 6. We already have two different answers, which means that this statement is not sufficient.(2) n is a multiple of 12. Every multiple of 12 is also a multiple of 6, thus n divided by 6 yields the remainder of 0. Sufficient.Answer:x^2+y^2. Sufficient.Answer:(2) p is the sum of the squares of two positive integers.Sol：(1) When p is divided by 8, the remainder is 5 p= 8q+5 = (8q+4) + 1 = 4(2q+1) + 1 so the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.(2) p is the sum of the squares of two positive integers since p is an odd integer then one of the integers must be even and another odd: p=(2n)^2 + (2m+1)^2 = 4n^2 + 4m^2 + 4m + 1 = 4(n^2+m^2+m) + 1the same way as above: the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.Answer:s check for several numbers which are not divisible by 2 or 3:n=1 n^21=0 remainder 0。 (4, 6)。s not n, hence either n1 or n+1).(1)+(2) From (1) (n1)(n+1) is divisible by 8,fr