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Cyclicity on the GMAT
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10 Feb 2016, 11:48
Cyclicity of Units Digits on the GMAT The first thing you need to understand is that when we multiply two integers together, the last digit of the result depends only on the last digits of the two integers. For example: 24 * 12 = 288 Note here: …4 * …2 = …8 So when we are looking at the units digit of the result of an integer raised to a certain exponent, all we need to worry about is the units digit of the integer. Let’s look at the pattern when the units digit of a number is 2. Units digit 2:2^1 = 22^2 = 42^3 = 82^4 = 1 62^5 = 3 22^6 = 6 42^7 = 12 82^8 = 25 62^9 = 51 22^10 = 102 4… Note the units digits. Do you see a pattern? 2, 4, 8, 6, 2, 4, 8, 6, 2, 4 … and so on So what will \(2^{11}\) end with? The pattern tells us that two full cycles of 2486 will take us to 2^8, and then a new cycle starts at 2^9. 2486 2486 24 The next digit in the pattern will be 8, which will belong to \(2^{11}\). In fact, any integer that ends with 2 and is raised to the power 11 will end in 8 because the last digit will depend only on the last digit of the base. So \(652^{11}\) will end in \(8,1896782^{11}\) will end in 8, and so on… A similar pattern exists for all units digits. Let’s find out what the pattern is for the rest of the 9 digits. Units digit 3:3^1 = 33^2 = 93^3 = 2 73^4 = 8 13^5 = 24 33^6 = 72 9The pattern here is 3, 9, 7, 1, 3, 9, 7, 1, and so on… Units digit 4:4^1 = 44^2 = 1 64^3 = 6 44^4 = 25 6The pattern here is 4, 6, 4, 6, 4, 6, and so on… Integers ending in digits 0, 1, 5 or 6 have the same units digit (0, 1, 5 or 6 respectively), whatever the positive integer exponent. That is: \(1545^{23} = ……..5\) \(1650^{19} = ……..0\) \(161^{28} = ………1\) Hope you get the point. Units digit 7:7^1 = 77^2 = 4 97^3 = 34 37^4 = …. 1 (Just multiply the last digit of 343 i.e. 3 by another 7 and you get 21 and hence 1 as the units digit) 7^5 = …. 7 (Now multiply 1 from above by 7 to get 7 as the units digit) 7^6 = …. 9The pattern here is 7, 9, 3, 1, 7, 9, 3, 1, and so on… Units digit 8:8^1 = 88^2 = 6 48^3 = … 28^4 = … 68^5 = … 88^6 = … 4The pattern here is 8, 4, 2, 6, 8, 4, 2, 6, and so on… Units digit 9: 9^1 = 99^2 = 8 19^3 = 72 99^4 = … 1The pattern here is 9, 1, 9, 1, 9, 1, and so on… Summing it all up:1) Digits 2, 3, 7 and 8 have a cyclicity of 4; i.e. the units digit repeats itself every 4 digits.Cyclicity of 2: 2, 4, 8, 6 Cyclicity of 3: 3, 9, 7, 1 Cyclicity of 7: 7, 9, 3, 1 Cyclicity of 8: 8, 4, 2, 6 2) Digits 4 and 9 have a cyclicity of 2; i.e. the units digit repeats itself every 2 digits.Cyclicity of 4: 4, 6 Cyclicity of 9: 9, 1 3) Digits 0, 1, 5 and 6 have a cyclicity of 1.Cyclicity of 0: 0 Cyclicity of 1: 1 Cyclicity of 5: 5 Cyclicity of 6: 6
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Veritas Prep GMAT Instructor
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Location: Pune, India

Re: Cyclicity on the GMAT
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10 Feb 2016, 11:55
Cyclicity in GMAT Remainder Questions Usually, cyclicity cannot help us when dealing with remainders, but in some cases it can. Today we will look at the cases in which it can, and we will see why it helps us in these cases. First let’s look at a pattern: 20/10 gives us a remainder of 0 (as 20 is exactly divisible by 10) 21/10 gives a remainder of 1 22/10 gives a remainder of 2 23/10 gives a remainder of 3 24/10 gives a remainder of 4 25/10 gives a remainder of 5 and so on… In the case of this pattern, 20 is the closest multiple of 10 that goes completely into all these numbers and you are left with the units digit as the remainder. Whenever you divide a number by 10, the units digit will be the remainder. Of course, if the units digit of a number is 0, the remainder will be 0 and that number will be divisible by 10 — but we already know that. So remainder when 467,639 is divided by 10 is 9. The remainder when 100,238 is divided by 10 is 8 and so on… Along the same lines, we also know that every number that ends in 0 or 5 is a multiple of 5 and every multiple of 5 must end in either 0 or 5. So if the units digit of a number is 1, it gives a remainder of 1 when divided by 5. If the units digit of a number is 2, it gives a remainder of 2 when divided by 5. If the units digit of a number is 6, it gives a remainder of 1 when divided by 5 (as it is 1 more than the previous multiple of 5). With this in mind: 20/5 gives a remainder of 0 (as 20 is exactly divisible by 5) 21/5 gives a remainder of 1 22/5 gives a remainder of 2 23/5 gives a remainder of 3 24/5 gives a remainder of 4 25/5 gives a remainder of 0 (as 25 is exactly divisible by 5) 26/5 gives a remainder of 1 27/5 gives a remainder of 2 28/5 gives a remainder of 3 29/5 gives a remainder of 4 30/5 gives a remainder of 0 (as 30 is exactly divisible by 5) and so on… So the units digit is all that matters when trying to get the remainder of a division by 5 or by 10. Let’s take a few questions now: What is the remainder when \(86^{183}\) is divided by 10?Here, we need to find the last digit of \(86^{183}\) to get the remainder. Whenever the units digit is 6, it remains 6 no matter what the positive integer exponent is. So the units digit of \(86^{183}\) will be 6. So when we divide this by 10, the remainder will also be 6. Next question: What is the remainder when \(487^{191}\) is divided by 5?Again, when considering division by 5, the units digit can help us. The units digit of 487 is 7. 7 has a cyclicity of 7, 9, 3, 1. Divide 191 by 4 to get a quotient of 47 and a remainder of 3. This means that we will have 47 full cycles of “7, 9, 3, 1” and then a new cycle will start and continue until the third term. 7, 9, 3, 1 7, 9, 3, 1 7, 9, 3, 1 7, 9, 3, 1 … 7, 9, 3 So the units digit of \(487^{191}\) is 3, and the number would look something like ……………..3 As discussed, the number ……………..0 would be divisible by 5 and ……………..3 would be 3 more, so it will also give a remainder of 3 when divided by 5. Therefore, the remainder of \(487^{191}\) divided by 5 is 3. Last question: If x is a positive integer, what is the remainder when \(488^{6x}\) is divided by 2?Take a minute to review the question first. If you start by analyzing the expression \(488^{6x}\), you will waste a lot of time. This is a trick question! The divisor is 2, and we know that every even number is divisible by 2, and every odd number gives a remainder 1 when divided by 2. Therefore, we just need to determine whether \(488^{6x}\) is odd or even. \(488^{6x}\) will be even no matter what x is (as long as it is a positive integer), because 488 is even and we know even*even*even……(any number of terms) = even. So \(488^{6x}\) is even and will give remainder 0 when it is divided by 2.
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Re: Cyclicity on the GMAT
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10 Feb 2016, 12:08
Cyclicity in GMAT Remainder Questions If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 In this problem, we are looking for the remainder when the divisor is 5. We know from last week that if we get the last digit of the dividend, we will be able to find the remainder, so let’s focus on finding the units digit of 3^(8n + 3) + 2. The units digit of 3 in a positive integer power has a cyclicity of: 3, 9, 7, 1 So the units digit of 3^(8n + 3) = 3^(4*2n + 3) will have 2n full cycles of 3, 9, 7, 1 and then a new cycle will start: 3, 9, 7, 1 3, 9, 7, 1 3, 9, 7, 1 … 3, 9, 7, 1 3, 9, 7 Since the exponent a remainder of 3, the new cycle ends at 3, 9, 7. Therefore, the units digit of 3^(8n + 3) is 7. When you add another 2 to this expression, the units digit becomes 7+2 = 9. This means the units digit of 3^(8n+3) + 2 is 9. When we divide this by 5, the remainder will be 4, therefore, our answer is E. This question is discussed HERE. Not so bad; let’s try a data sufficiency problem: If k is a positive integer, what is the remainder when 2^k is divided by 10?Statement 1: k is divisible by 10 Statement 2: k is divisible by 4 With this problem, we know that the remainder of a division by 10 can be easily obtained by getting the units digit of the number. Let’s try to find the units digit of 2^k. The cyclicity of 2 is: 2, 4, 8, 6. Depending on the value of k is, the units digit of 2^k will change: If k is a multiple of 4, it will end after one cycle and hence the units digit will be 6. If k is 1 more than a multiple of 4, it will start a new cycle and the units digit of 2^k will be 2. If k is 2 more than a multiple of 4, it will be second digit of a new cycle, and the units digit of 2^k will be 4. If k is 3 more than a multiple of 4, it will be the third digit of a new cycle and the units digit of 2^k will be 8. If k is 4 more than a multiple of 4, it will again be a multiple of 4 and will end a cycle. The units digit of 2^k will be 6 in this case. and so on… So what we really need to find out is whether k is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4, or three more than a multiple of 4. Statement 1: k is divisible by 10 With this statement, k could be 10 or 20 or 30 etc. In some cases, such as when k is 10 or 30, k will be two more than a multiple of 4. In other cases, such as when k is 20 or 40, k will be a multiple of 4. So for different values of k, the units digit will be different and hence the remainder on division by 10 will take multiple values. This statement alone is not sufficient. Statement 2: k is divisible by 4 This statement tells you directly that k is divisible by 4. This means that the last digit of 2^k is 6, so when divided by 10, it will give a remainder of 6. This statement alone is sufficient. therefore our answer is B. This question is discussed HERE. Now, to cap it all off, we will look at one final question. It is debatable whether it is within the scope of the GMAT but it is based on the same concepts and is a great exercise for intellectual purposes. You are free to ignore it if you are short on time or would not like to go an iota beyond the scope of the GMAT: What is the remainder of (3^7^11) divided by 5?(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 For this problem, we need the remainder of a division by 5, so our first step is to get the units digit of 3^7^{11}. Now this is the tricky part – it is 3 to the power of 7, which itself is to the power of 11. Let’s simplify this a bit; we need to find the units digit of 3^a such that a = 7^{11}. We know that 3 has a cyclicity of 3, 9, 7, 1. Therefore (similar to our last problem) to get the units digit of 3^a, we need to find whether a is a multiple of 4, one more than a multiple of 4, two more than a multiple of 4 or three more than a multiple of 4. We need a to equal 7^{11}, so first we need to find the remainder when a is divided by 4; i.e. when 7^{11} is divided by 4. For this, we need to use the binomial theorem we learned earlier in this post (or we can use the method of “pattern recognition”): The remainder of 7^{11} divided by 4 = The remainder of (4 + 3)^{11} divided by 4 = The remainder of 3^{11} divided by 4 = The remainder of 3*3^{10} divided by 4 = The remainder of 3*9^5 divided by 4 = The remainder of 3*(8+1)^5 divided by 4 = The remainder of 3*1^5 divided by 4 = The remainder of 3 divided by 4, which itself = 3 So when 7^{11} is divided by 4, the remainder is 3. This means 7^{11} is 3 more than a multiple of 4; i.e. a is 3 more than a multiple of 4. Now we go back to 3^a. We found that a is 3 more than a multiple of 4. So there will be full cycles (we don’t need to know the exact number of cycles) and then a new cycle with start with three digits remaining: 3, 9, 7, 1 3, 9, 7, 1 3, 9, 7, 1 … 3, 9, 7, 1 3, 9, 7 With this pattern, we see the last digit of 3^7^11 is 7. When this 7 is divided by 5, remainder will be 2 – therefore, our answer is C. This question is discussed HERE.
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Re: Cyclicity on the GMAT
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10 Feb 2016, 12:13



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Cyclicity on the GMAT
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20 Jul 2018, 18:31
KarishmaB Bunuel niks18 chetan2uQuote: So what will \(2^{11}\) end with? The pattern tells us that two full cycles of 2486 will take us to 2^8, and then a new cycle starts at 2^9.
2486
2486
24
The next digit in the pattern will be 8, which will belong to \(2^{11}\).
Do I need to write such a pattern every time? Can you please advise how to form correct multiples of exponent power? E.g. 11 could be written as (2*5) + 1 Also what do we consider if we have exponent power as 12 Since 12 can be written as 3* 4 or 4*3 ?
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Re: Cyclicity on the GMAT
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20 Jul 2018, 20:20
adkikani wrote: KarishmaB Bunuel niks18 chetan2uQuote: So what will \(2^{11}\) end with? The pattern tells us that two full cycles of 2486 will take us to 2^8, and then a new cycle starts at 2^9.
2486
2486
24
The next digit in the pattern will be 8, which will belong to \(2^{11}\).
Do I need to write such a pattern every time? Can you please advise how to form correct multiples of exponent power? E.g. 11 could be written as (2*5) + 1 Also what do we consider if we have exponent power as 12 Since 12 can be written as 3* 4 or 4*3 ? You have to get the exponent to nearest smaller multiple of 4.. A) so if it is 11, multiple of 4 just smaller than 11 is 8, so it becomes 4*2+3 So the units digit of exponent 11 will be same as that of power 3 B) you have to convert the power to multiple of 12 so it is always 4k, where k is an integer so 4*3, although 3*4 or 4*3 does not make difference till the time you work knowing you have multiple of 4 in exponent
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Re: Cyclicity on the GMAT &nbs
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20 Jul 2018, 20:20






