1、Introduction (赶时间的童鞋可以略过。 。 。只是一些概念帮助童鞋们回忆余数)DefinitionIf x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y= divisor * quotient + remainder = xq + r; and 0 n, what is the remainder when p2 - n2 is divided by 15 ?(1) The r
2、emainder 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 p2 - n2 just from this information.So, INSUFFICIENTSTAT2 : The remainder when p - n is divided by 3 is 1p-n
3、 = 3s + 1but we cannot say anything about p2 - n2 just from this information.So, INSUFFICIENTSTAT1+STAT2：p2 - n2 = (p+n) * (p-n) = (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 sa
4、y anything about the reminder as its value will change with the values of k and s.So INSUFFICIENTHence answer will be 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. Sol:r is the remainder when 4 + 7n i
5、s divided by 37n + 4 can we written as 6n + n + 3+ 1 = 3(2n+1) + n +1reminder of 7n+4 by 3 will be same as reminder of 3(2n+1) +n +1 by 33*(2n+1) will go by 3 so the reminder will be the same as the reminder of (n+1) by 3.STAT1: n+1 is divisible by 3n+1 = 3k (where k is an integer)n+1 will give 0 as
6、 the reminder when divided by 3so, 7n+4 will also give 0 as the reminder when its divided by 3 (as its reminder is same as the reminder for (n+1) when divided by 3 = r =0So, SUFFICIENTSTAT2: n20.we cannot do anything by this information as there are many values of nso, INSUFFICIENT.Hence, answer wil
7、l be APractice: If x is an integer, is x between 27 and 54?(1) The remainder when x is divided by 7 is 2.(2) The remainder when x is divided by 3 is 2. 再次请大家来考验下自己吧O(_)OSol: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 t
8、he values of xso, INSUFFICIENTSTAT2: The remainder when x is divided by 3 is 2. x = 3s + 2Possible values of x are 2,5,8,11,.,53,.we cannot say anything about the values of xso, INSUFFICIENTSTAT1+STAT2：now there are two approaches1. write the values of t from stat1 and then from stat2 and then take
9、the common valuesfrom STAT1 x = 2,9,16,23,30,37,44,51,58,.,65,.from STAT2 x = 2,5,8,.,23,.,44,.,59,65,.common values are x = 2,23,44,65,.2. equate x = 7k+2 to x=3s+2we have 7k + 2 = 3s+2k = 3s/7since, k is an integer so only those values of s which are multiple of 7 will satisfy both STAT1 and STAT2
10、so, common values are given by x = 3s + 2 where s is multiple of 7so x = 2,23,44,65 (for s=0,7,14,21 respectively)Clearly there are values of x which are between 27 and 54 (i.e. 44) and those which are not (2,23,65)So, both together also INSUFFICIENTSo, answer will be E方法三：MOD 法请大家参见知之为之之大侠的帖子！地址：2.
11、 Remainder QuestionPatternsBackgroundl Most GMAT remainder problems are encountered in data sufficiency section.l All GMAT remainder questions are limited to positive integers only.l Both number plugging method and algebra are suitable tosolve remainder questions.l Some remainder questions can be di
12、sguised as word problems. See below.l Usually you get 1, maximum 2 questions on remainders on thetest (based on GMAT Prep CATs)以下 Pattern 并不是按照重要性顺序来排的哟Pattern 6 is the most common pattern! 妹纸觉得大家都可以看一看。 。 。当做练习吧Pattern#1: The ratio of two integers is given and we are asked to find possible value of
13、 the remainder when one integer is divided by another.Q1： OG13 diagnostic test, question 13If s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t ?(A) 2(B) 4 (C) 8 (D) 20 (E) 45Sol：S divided by t yields the remainder of r can alwa
14、ys be expressed as: s/t = q + r/t (which is the same as s= qt+r), where q is the quotient and r is the remainder.Given that s/t = 64.12 = 64(12/100) = 64(3/25) = 64 +3/25,so according to the above r/t=3/25, which means that r must be a multiple of 3. Only option E offers answer which is a multiple o
15、f 3Answer. E.Q2： OG13 Practice Questions, question 95When positive integer x is divided by positive integer y,the remainder is 9. If x/y = 96.12, what is the value of y?(A) 96(B) 75(C) 48(D) 25(E) 12Sol：When positive integer x is divided by positive integer y,the remainder is 9 x=qy+9;x/y=96.12 x=96
16、y+0.12y (so q above equals to 96);0.12y=9 y=75.Answer: Bfinding the remainder when an expression with variable is divided by some integer.OG13 Practice Questions, question 26If n is a prime number greater than 3, what is the remainder when n2 is divided by 12 ?(A) 0(B) 1(C) 2(D) 3(E) 5Sol：There are
17、several algebraic ways to solve this question, but the easiest way is as follows: since we cannot have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12.n=5 n2=25 remainder upon division 25 by 12 is 1.Answer: B.Pattern#3: mi
18、n/max question involving remaindersWhen positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35.A. 3B. 4C. 12D. 32E. 35Sol：Positive integer n is divided by 5, the remainder is 1 n
19、= 5q+1, where q is the quotient 1, 6, 11, 16,21, 26,31, .Positive integer n is divided by 7, the remainder is 3 n=7p+3, where p is the quotient 3, 10, 17, 24, 31,You cannot use the same variable for quotients in both formulas, because quotient may not be the same upon division n by two different num
20、bers.For example 31/5, quotient q=6 but 31/7, quotient p=4.There is a way to derive general formula for n (of a type n= mx+r, where x is divisor and r is a remainder)based on above two statements: Divisor x would be the least common multiple of above two divisors 5 and 7, hence x=35.Remainder r woul
21、d be the first common integer in above two patterns, hence r=31.Therefore general formula based on both statements is n=35m+31.Thus 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: BPattern#4: disguised PS remainder problem.There are between 100
22、 and 110 cards in a collection of cards. If they are counted out 3 at a time, there are 2 left over, but if they are counted out 4 at a time, there is 1 left over. How many cards are in the collection?(A) 101(B) 103(C) 106(D) 107(E) 109Sol：If the cards are counted out 3 at a time, there are 2 leftov
23、er: x=3q+2. From the numbers from 100 to 110 following three give the remainder of 2 upon division by 3: 101, 104 and 107;If the cards are counted out 4 at a time, there are 1 leftover: x=4p+1. From the numbers from 100 to 110 following three give the remainder of 1 upon division by 4: 101, 105 and
24、109. Since x, the number of cards, should satisfy both conditions then it equals to 101.Answer:APattern#5: we need to answer some question about an integer, when the statements give info involving remainders.Q1： OG13 Practice Questions, question 58What is the tens digit of positive integer x ?(1) x
25、divided by 100 has a remainder of 30.(2) x divided by 110 has a remainder of 30.Sol：(1) x divided by 100 has a remainder of 30 x =100q+30, so x can be: 30, 130, 230, . Each has the tens digit of 3. Sufficient.(2) x divided by 110 has a remainder of 30 x =110p+30, so x can be: 30, 250, . We already h
26、ave two values for the tens digit. Not sufficient.Answer:A.Q2： OG13 Practice Questions, question 83If k is an integer such that 56 k is an odd number, thus it could be 57, 59, 61, 63, or 65. Not sufficient.(2) If k + 1 were divided by 3, the remainder would be 0 k is 1 less than a multiple of 3, thu
27、s it could be 59, 62, or 65. Not sufficient.(1)+(2) k could still take more than one value: 59 or 65. Not sufficient.Answer: E.Pattern#6: we need to find the remainder when some variable or an expression with variable(s) is divided by some integer. Usually the statements give divisibility/remainder
28、info. Most common 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 alr
29、eady 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: B.Q2： If x and y are integer, what is the remainder when x2 + y2 is divided by
30、 5? (1) When x-y is divided by 5, the remainder is 1(2) When x+y is divided by 5, the remainder is 2Sol：(1) When x-y is divided by 5, the remainder is 1 x-y = 5q+1, so x-y can be 1, 6, 11, . Now, x=2 and y=1 (x-y=1) then x2+y2= 5 and thus the remainder is 0, but if x=3 and y=2 (x-y=1)then x2+y2= 13
31、and thus the remainder is 3. Not sufficient.(2) When x+y is divided by 5, the remainder is 2 x+y=5p+2, so x+y can be 2, 7, 12, . Now, x=1 and y=1 (x+y=2) then x2+y2= 2 and thus the remainder is 2, but if x=5 and y=2 (x+y=7) then x2+y2= 29 and thus the remainder is 4. Not sufficient.(1)+(2) Square bo
32、th expressions: x2 -2xy +y2= 25q2 + 10q+ 1 and x2 +2xy +y2= 25p2 + 20p + 4 add them up: 2(x2+y2)= 5(5q2+2q+5p2+4p+1) so 2(x2+y2) is divisible by 5 (remainder 0), which means that so is x2+y2. Sufficient.Answer: C.Q3： If t is a positive integer and r is the remainder when t2+5t+6 is divided by 7, wha
33、t is the value of r?(1) When t is divided by 7, the remainder is 6.(2) When t2 is divided by 7, the remainder is 1.Sol：First of all factor t2+5t+6 t2+5t+6=(t+2)(t+3)(1) When t is divided by 7, the remainder is 6 t=7q+6 (t+2)(t+3)=(7q+8)(7q+9). Now, no need to expand and multiply all the terms, just
34、notice that when we expand all terms but the last one, which will be 8*9=72, will have 7 as a factor and 72 yields the remainder of 2 upon division by 7. Sufficient. (2) When t2 is divided by 7, the remainder is 1 different values of t possible: for example t=1 or t=6, which when substituted in (t+2
35、)(t+3) will yield different remainder upon division by 7. Not sufficient.Answer: A.Q4： If 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：(1) When p is divided by
36、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
37、 of the integers must be even and another odd: p=(2n)2 + (2m+1)2 = 4n2 + 4m2 + 4m + 1 = 4(n2+m2+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: D.Q5： If n is a po
38、sitive 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 3Sol：Plug-in method:(n-1)(n+1) = n2-1(1) n is not divisible by 2 pick two odd numbers:lets say 1 and 3 if n=1, then n2-1=0 and as zero is divisible by 2
39、4 (zero is divisible by any integer except zero itself) so remainder is 0 but if n=3, then n2-1=8 and 8 divided by 24 yields remainder of 8. Two different answers, hence not sufficient.(2) n is not divisible by 3 pick two numbers which are not divisible by 3: lets say 1 and 2 if n=1, then n2-1=0,so
40、remainder is 0 but if n=2, then n2-1=3 and 3 divided by 24 yields remainder of 3. Two different answers, hence not sufficient.(1)+(2) Lets check for several numbers which are not divisible by 2 or 3:n=1 n2-1=0 remainder 0;n=5 n2-1=24 remainder 0;n=7 n2-1=48 remainder 0;n=11 n2-1=120 remainder 0.Well
41、 it seems that all appropriate numbers will give remainder of 0. Sufficient.Algebraic approach:(1) n is not divisible by 2. Insufficient on its own, but this statement says that n=odd n-1 and n+1 are consecutive even integers (n-1)(n+1) must be divisible by 8(as both multiples are even and one of th
42、em will be divisible by 4. From consecutive even integers one is divisible by 4: (2, 4); (4, 6); (6, 8); (8,10); (10, 12), .).(2) n is not divisible by 3. Insufficient on its own, but form this statement either n-1 or n+1 must be divisible by 3 (as n-1, n, and n+1 are consecutive integers, so one of
43、 them must be divisible by 3, we are told that its not n, hence either n-1 or n+1).(1)+(2) From (1) (n-1)(n+1) is divisible by 8,from (2) its also divisible by 3, therefore it must be divisible by 8*3=24,which means that remainder upon division (n-1)(n+1) by 24 will be 0. Sufficient.Answer: C.Patter
44、n#7: disguised DS remainder problem.A person inherited few gold coins from his father. If he put 9 coins in each bag then 7 coins are left over. However if he puts 7 coins in each bag then 3 coins are left over. What is the number of coins he inherited from his father?(1) The number of coins lies be
45、tween 50 to 120.(2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200.Sol：If he puts 9 coins in each bag then 7 coins are left over c=9q+7, so # of coins can be: 7, 16, 25, 34, 43, 52, 61, .If he puts 7 coins in each bag then 3 coins are left over c=7p
46、+3, so # of coins can be: 3, 10, 17, 24, 31, 38, 45,52, 59, .General formula for c based on above two statements will be: c=63k+52 (the divisor should be the least common multiple of above two divisors 9 and 7, so 63 and the remainder should be the first common integer in above two patterns, hence 5
47、2). C=63k+52 means that # of coins can be: 52, 115, 178,241, .(1) The number of coins lies between 50 to 120 # of coins can be 52 or 115. Not sufficient.(2) If he put 13 coins in one bag then no coin is left over and number of coins being lesser than 200 # of coins is a multiple of 13 and less than
48、200: only 52 satisfies this condition. Sufficient.Answer: B.Pattern#8: C-Trap remainder problem. “C trap“ is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.If a and b are po
49、sitive integers, what is the remainder when 4(2a+1+b) is divided by 10?(1) a = 1(2) b = 2Sol：4 in positive integer 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 4x/10 we should know whether the power is odd or even: if its odd the remainder will be 4 and if its 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)= 4odd the re
