Cyclicity of last two digits:
Nowadays, in some of the competitive MBA entrance examinations, questions have been asked to find out the last two digits of a numberany power.
There are basically three ways to find out the last two digits of a number.
a) Binomial Theorem: We can find last two digits of number ending with specific unit’s digit.
b) Remainder: We have already seen this method of finding out the last two digits of a number in REMAINDER chapter. We can use this method on all the numbers ending with a unit’s place of 2, 4, 6, 8 or 5.
c) Cyclicity: And third method is to use the cyclicity of last two digits. This method can also be used with specific digits.
We should be well-conversed with all the three methods as at some places Binomial Theorem comes in handy, while at other places remainder method or Cyclicity method will be beneficial. So, we choose either of the methods depending on which one of them will solve our purpose quickly.
Binomial Theorem:
Before learning Binomial Expansion, we will revisit some of the basic formulas encountered in school.
(a + b)2 = a2 + 2ab + b2.
(a + b)3 = a3 + 3a2b + 3ab2 + b3.
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.
So, for any expansion of the type (a + b)x, there will be ax, bx and some middle terms involving a & b with different coefficients . Same logic is also used for the Binomial Expression, only difference is that in these two formulas written above, we know the power. But, in case of binomial expansion, we are trying to generalize for any powers which we wish to take. And to take care of coefficients, nC0, nC1 and other terms will be used. All these terms have their origin in Permutation & Combination which will be discussed later. But, for the meanwhile you can just use them as formula.
Using this, we can find value of any such term.
Same logic is used for the expansion of (a + b)n.
(a +b)n means (a + b) (a + b) (a + b) (a + b)…..multiplied ‘n’ times. When (a + b) is multiplied twice, we get a2 and b2 for sure and remaining term will be product of a & b. Let us see the expansion.
The formula can be learnt very easily if we observe the pattern carefully. The sum of the powers (n + 0) of a and b (of any term) is always n. So, we can conclude that the sum of the powers will always be the power given in the question.
Also, the value of nC0 = 1 = nCn. (nC0 and nCn are the coefficients of the first and last terms of expansion).
The value of nC1 = nCn –1 = n. (nC1 and nCn –1 are the coefficients of the second and the second last terms of expansion).
Similarly, we can find the values for other coefficients, but they won’t be needed.
By observing carefully, we can deduce that coefficient of the first term is equal to the coefficient of the last term. Similarly, the coefficient of the second term is equal to that of second last term. And so on.
Even when there is a negative sign in the expansion, the formula will remain the same except the negative sign in alternate terms.
For example, we want to write the expansion of (a – b)n.
The values of the coefficients will be same as we calculated earlier.
After learning these two expansions, we will see their applications and the ease with which the last two digits of a number can be found.
Most of the students are scared of the formula and they don’t even make an effort to understand it logically. If you grasp it once, the sums can be solve in two or three steps. We can use Binomial Theorem where a number can be written as the sum of two numbers or difference of two numbers with one of the number having 0 at the unit’ s place.
For e.g. 11 = (1 + 10), 79 = (80 – 1), or 132 = 169 which can be written as (– 1 + 170) or 172 = 289 = (– 1+ 290) . So,the trick is to identify places where Binomial Theorem can be used efficiently. So, when any number can be represented in the form of addition or subtraction of two numbers with one number ending with a zero at the unit’s place, binomial expansion can be used.
Point Worth Noting: In expansion of (a + b)2: total of 3 terms.
In expansion of (a + b)3: total of 4 terms.
In expansion of (a + b)4: total of 5 terms.
So, in expansion of (a + b)n: total of (n + 1) terms.
Applications Of Binomial Expansion:
E.g. 1: Find the coefficient of x3 in the product of (x – 3) (x + 5) (x – 1 ) (x + 2) (x – 8).
To obtain all the terms involving x3, we need to take x from any three brackets and constants from the remaining two brackets. And after making all the possibilities, we will have to add up all of them which will give us the final answer.
If we take x from the first three brackets, we will have to take constant from the other two. And their product is – 16x3.
If we take x from the first two brackets and another x from the fourth one, then we will have to take constants from the third and the fifth bracket, and the product is 8x3.
Or in other words, we can say that the coefficient is equal to the sum of the product of the constants –3, 5, –1, 2, –8 taken two at a time.
Thus the required coefficient is = –15 + 3 –6 + 24 –5 + 10 –40 –2 + 8 –16 = –39.
E.g. 2: Find the expansion of (x + y)6.
We can say that in expansion of (x + y)6, there will be 7 terms of which two terms are x6 and y6. Remaining five terms will be of combination of x and y with the sum of powers of (x & y) be equal to 6.
On substituting values of the coefficients after calculating them, we get
E.g. 3: Find the coefficient of x16 in the expansion of (x2 – 2x)10.
E.g. 4: Find the fourteenth term of (2 – b)15.
When we expand this expression, we will get a total of 16 terms.
So, 16th term will be the last term with coefficients 15C15, the coefficients of 15th term will be 15C14 and 14th term will be with coefficients 15C13.
And the remaining terms of 14th term will be 22 × (– b)13. Remember the sum of powers of 2 and b has to 15.
So, the 14th term will be 15C13 × 22 × (– b)13 = 15C2 × (– 4b13) = – 460b13. (Remember value of 15C13 will be equal to 15C2).
E.g. 5: Find the two right-most digits of 9145.
We can use Binomial expansion here.
We have to find the last two digits. Starting from the third term onwards, the last two digits will have two zeros or more than two zeros as the power of 90 is increasing. Thus the last two digits of 9145 will be dependent on the first two terms of the expansion as all the terms ahead of it will have two zeros or more than two zeros which will not have an effect on the last two digits.
For e.g., if we want to find the last two digits of this series: 15 + 120 + 1400 + 1000 + 120000 + 23980000. Answer will be dependent on first two terms as all the terms ahead of it has got 2 or more than 2 zeros.
So, coming back to the previous question, we can conclude that the last two digits of 9145 will be dependent on first two terms of expansion.
So, the answer will be = 1 + 45 × 1 × 90 = 1 + 50 = 51. (45C0 = 1 and 45C1 = 45)
The above used method can also be used to find last three or last four digits, and in those cases answer will just depend on the first three and first four terms respectively.
E.g. 6: Find the two right-most digits of 8995.
.
Again, starting from the third term onwards we will start getting two zeros or more than two zeros at the end of that number. So, answer is not dependent on those terms, we just have to concentrate on the first two terms. Also, don’t worry about the negative number since we will get negative because of (-1)93 or (-1)91 or so on. Since, 8995 will be a positive number we don’t need to be bothered about the negative part.
So, finally answer will be: -1 + 95 × 1 × 90 = – 1 + 50 = 49. (95C0 = 1 and 95C1 = 95)
E.g. 7: Find the two right-most digits of 9364.
We learnt in the previous two questions that if a number can be represented in the form of (1 + something) or (-1 + something), we can use the binomial expansion very conveniently.
We can use that logic here also by manipulating the question given according to our need. Any number ending with 3 or 7 at the unit’s place when squared gives 9 at the unit’s place. If we get 9 at the unit’s place, we can write it in the format of (- 1 + something).
Let’s see how it works.
93 when squared gives us 8649, we can write 9364 as 864932.
higher powers of 8650 having more than 2 zeros.
Using the logic learnt in the previous examples, we can say that the right-most two digits will be 1 – 32 × 8650.
Last two digits will be 01 – 00 = – 99
What is the meaning of a number ending with – 99?
Remaining terms of expansion of (–1 + 8650)32 will be a positive number since 864932 is a positive number.
Thus, – 99 + a very large positive number ending with more than 2 zeroes.
So, the last two digits will be 01. ( for e.g., –99 + 1000000 = 01 at the last two digits)
E.g. 8: Find the two right-most digits of 3775.
Again square of a number having 7 at the unit’s place when squared results in 9 at the unit’s place.
So, we can use the logic of the previous questions.
3775 = 37 × 3774 = 37 × 136937.
Higher powers of 1370.
So, answer will be just dependent on the first two terms : – 1 + 37 × 1370 = – 1 + 90 = 89.
So, the last two digits of 136937 is 89, but the question is asking to find the last two digits of 37 × 136937.
So, finally the last two digits of 37 × 136937 = 37 × 89 = 93.
Alternative Method of finding out last two digits:
To find the last two digits, just divide by 100 and whatever is the remainder will be the last two digits of that number. We have already discussed this part in Remainders (Composite Index) This method is useful when we have to find the last two digits of a number which has a common factor related with 2, 4, 6, 8 or 5. For e.g. if we want to find the last two digits of 462 or 7860 or 122100 or 9534, we will divide this by 100. And the dividend and the divisor have something in common which can be cancelled, remainder can be calculated pretty easily by making cycles. We have already discussed this in detail in Remainder chapter. There is one more method for all such numbers. We can make cycles of last two digits of all such numbers as they start repeating itself. With the help of cycle, we can determine the last two digits. For these digits, cycle is pretty short. But, we should know how to multiply efficiently.
If we want to find out the last two digits of 782, traditional way is to multiply entire number. But that process when done multiple times will consume a lot of time. So, we need to look out for some smarter approach.
When we multiply two numbers and want to find the last two digits, we just need to find the last two digits and that will give us the required number.
E.g. 9: Find the last two digits of 4494.
We will find the cycle of the right-most two digits of the powers of 44.
After this the cycle will repeat itself. We can observe the cycle of the last two digits of 4494 is 44, 36, 84, 96, 24, 56, 64, 16, 04, and 76. A cycle of 10 terms are repeating. So, the last two digits of 4494 will be the 4th term (we divide 94 ÷ 10, we get 10 cycles complete & 4 remainder) in the cycle which is 96.
Answer is 96.
E.g. 10: Find the last two digits of 56173.
After this term, the last two digits will repeat. The cycle of the last two digits of 56173 is 56, 36, 16, 96, 76. So, the cycle is of 5.
Thus the last two digits of 56173 will be (173 ÷ 5 = some cycles and 3 remainder) the 3rd term of the cycle which is 16.
E.g. 11: Find the last two digits of 5597.
Thus, the cycle of the last two digits of 5597 is 55, 25, 75, 25, 75,……….
All the even powers of 55 will result in 25 at the last two places and all the even powers of 55 will end with 75 (except the first power which is 55).
So, the answer is 75.
Exercise:
1. Find the two right-most digits of 8476.
2. Find the two right-most digits of 10395.
3. Find the two right-most digits of 4199.
4. Find the two right-most digits of 4793.
5. Find the two right-most digits of 97144.
6. Find the two right-most digits of 7577.
7. Find the last two digits of 62000.
8. Find the digit at the ten’s place of 119123.
9. Find the right-most two digits of 7405.
10. Find the right-most two digits of 6258.
11. Find the 28th term of (3x + y)30.
12. Find the 12th term of (2x – 1)13.
Answer Key:
1. 16 2. 07 3. 61 4. 27 5. 81 6. 75
7. 76 8. 5 9. 07 10. 04
11. 109620(x3y27) 12. [–312x2]
Explanations:
Question 1:
We can find out the last two digits of 8476 using two methods.
a) First method: Since 84 has a common factor of 4. So we can divide this number by 100 and find out the last two digits by cancelling out the common terms.
b) We can frame cycles of 8476 to find the last two digits.
The last two digits of 8476 do have a cycle. The cycle goes like 84, 56, 04, 36, 24, 16, 44, 96, 64, 76,……………….. So, the last two digits of 8476 have cycle of 10.
So, 8476 will be the 6th term of the cycle which is 16.
Question 2:
Two right-most digits of 10395 can be found out using binomial theorem as the unit’s place of 10395 on squaring will end up with 9.
On squaring, 10395 = [(103)2]47 × 103 = (10609)47 × 103.
Last two digits of 1060947 can be found out by writing it as (–1 + 10610)47. Now, we can apply the binomial theorem and we know in all such cases, last two digits ids dependent on the first two terms.
So, (–1 + 10610)47= (–1)47 + 47 × (–1)46 × 10610 + terms having two zeros or more.
On solving, we get = – 1+ 47 × 10610 = – 1 + 70 = 69 as the last two digits.
But the question is asking to find last two digits of (10609)47 × 103 = 69 × 103 = 07 is the answer.
Question 3:
On applying binomial theorem, 4199 = (1 + 40)99 = 199 + 99 × 198 × 40 + terms having two zeros or more.
On simplifying, we get = 1 + 60 = 61 is the last two digits of 4199.
Question 4:
4793 = [(47)2]46 × 47 = (2209)46 × 47.
We can find last two digits of (2209)46 using Binomial theorem.
(–1 + 2210)46 = (–1)46 + 46 × (–1)45 × 2210 + terms having two zeros or more.
(–1 + 2210)46 = 1 – 46 × 2210 + terms having two zeros or more.
On simplifying, we get 1 – 60 + terms having two zeros or more = – 59 + terms having two zeros are more.
So, the last two digits of (2209)46 is 41 as [– 59 added to a term having two zero or more will end up with 41 at the last two digits].
But, the question is to find last two digits of (2209)46 × 47 which is equal to 41× 47 = 27 as the last two digits.
Question 5:
97144 can be written as [(97)2]72 = (9409)72 = (– 1 + 9410)72.
(– 1 + 9410)72 = (– 1)72 + 72 × (– 1)71 × 9410 + terms having two or more zeros.
On solving, we get = 1 – 72 × 9410 + terms having two or more zeros = 1 – 20 + terms having two or more zeros = – 19 + terms having two or more zeros = 81 as the last two digits.
So, answer is 81.
Question 6:
Last two digits of 7577 can be found out by framing cycles.
So, last two digits of 7577 are following cycle of 75, 25, 75, 25 and so on. All the odd powers of 7577 will end up with last two digits of 75.
Question 7:
Last two digits of 62000 can be obtained by framing the cycles.
So, the cycles of 62000 is like 06, 36, 16, 96, 76, 56, 36, 16, ………. The cycle is repeating from the second term onwards.
36 will be 2nd term, 7th term, 12th term and so on.
Similarly, 76 will be the 5th term, 10th term, 15th term, 20th term, 25th term and so on. That means all the terms which are multiple of 5 will end up with last two digits of 76.
So, last two digits of 62000 should be 76.
Question 8:
119123 can be written as = (– 1 + 120)123, now we can apply binomial theorem.
(– 1 + 120)123 = (– 1)123 + 123 × (–1)122 × 120 + terms having two or more zeros.
On simplifying, we get = – 1 + 123 × 120 + terms having two or more zeros = – 1 + 60 = 59.
So, last two digits of 119123 = 59 and the digit at the ten’s place is 5.
Question 9:
The last two digits of 7405 can be found by squaring it and converting the number ending with 9 at the unit’s place.
So, 7405 = [(7)2]202 × 7 = (49)202 × 7.
We can use binomial theorem to find the last two digits of 49202.
49202 = (– 1 + 50)202 = (– 1)202 + 202 × (– 1)201 × 50 + terms having two or more than two zeros.
On simplifying, we get 1 – 202 × 50 + terms having two or more than two zeros = 1 – 10100 + terms having two or more than two zeros = –99 + terms having two or more than two zeros.
So, the last two digits of 49202 are 01.
But, the question is asking to find the last two digits of (49)202 × 7 which will (01) × 7 = 07.
Question 10:
To find the last two digits of 6258, we will have to make cycles.
The cycle of 6258 will repeat after 6221.
And the cycle goes like 62, 44, 28, 36, 32, 84, 08, 96, 52, 24, 88, 56, 72, 64, 68, 16, 92, 04, 48, 76, 12,……..
So, 44 will be the 2nd term, 22nd term, 42nd term and so on.
Similarly, 04 will be the 18th term, 38th term, 58th term and so on.
And in the question, we need to find 6258 which is 04.
11. In the expansion of (3x + y)30, there will be 31 terms.
Coefficient of 31st term will be 30C30
Coefficient of 30th term will be 30C29
Coefficient of 29th term will be 30C28
Coefficient of 28th term will be 30C27.
And the 28th term will be 30C27 × (3x)3 × y27.
So, the value of 28th term is 30C27 × (3x)3 × y27 = 140 × 29 × 27 × x3y27 = 109620(x3y27).
12. There will be 14 terms in total in expansion of (2x – 1)13.
Coefficient of 14th term will be 13C13.
Coefficient of 13th term will be 13C12.
Coefficient of 12th term will be 13C11.
So, the 12th term will be 13C11 × (2x)2 × (– 1)11. (Sum of the powers of the two terms will be always equal to 13)
Value of 13C11 is equal to 13C2.
So, value of 12th term will be [–312x2] as (the value of 13C2 = 78 )