These are the commonly found questions in CAT which are easy to solve but then, many aspirants make the mistake of falling into the trap of ‘overthinking’ and ‘oversolving’ these questions and make a mess of simple questions.
Finding the last digit of an expression
This concept is useful while not only solving straightforward questions but also across a variety of topics including the one that we had previously covered where we found the last non-zero digit of a factorial.
We simply use a concept called cyclicity of the unit place digit and find the solution easily.
The only thing that we have to understand here is that the unit’s place digit is the only thing that matters and so, we need not calculate the entire thing. A simple observation is enough to understand that there would be a trend when it comes to calculating unit’s place digits.
|Numbers ending with||Sequence of unit’s place digits||Numbers ending with||Sequence of unit’s place digits|
|2||2, 4, 8, 6||7||7, 9, 3, 1|
|3||3, 9, 7, 1||8||8, 4, 2, 6|
|4||4, 6||9||9, 1|
The things to understand from the above table are:
1) Every number follows a maximum cyclicity of 4. That is, a1, a5, a9, …a4k+1 will all have the same unit’s place digit irrespective of the nature of a and so will a2, a6, a10, …a4k+2 AND a3, a7, a11, …a4k+3 AND a4, a8, a12, …a4k. Once you understand and remember the above table and the cyclicity of powers, you should be good to find the last digit of any number in the form of ab within a few seconds.
Let’s solve a few examples
What is the unit’s place digit of 79080?
As 9080 is divisible by 4 and so, in the form of 4k, the unit’s place digit would be the same as that of 74 which is 1 as can be seen.
What is the unit’s place digit of 23412367?
As 12367 is in the form of 4k+3, we understand that the expression is in the form of 43 which ends in 4.
Understand that no other digits except that in the unit’s place will have any bearing on the eventual outcome and so, all the other digits in the base can be neglected.
Last 2 digits:
This is a bit trickier and students tend to get confused while solving these questions. While a few do manage to solve it using the a mod 100 approach, it does become tedious in a few cases. So, the better way would be to understand generic methods that lead us to the answer. There are essentially 3 cases that could be present:
Last digit of the base ends in 1, 3, 7 or 9
Step I: Bring it down to the form of (a4)k * (a smaller manageable number)
Step II: Now that the unit’s place digit will be 1, the only job left is to find the tens place digit. This can be found by multiplying the tens place digit of the base with the unit’s place digit of the power. Sounds confusing? Let’s see a couple of examples:
Find the last two digits of 72008 (CAT 2008)
The power is already in the form of 4k and so, we will simply write it as (74)502
We will consider only the last two digits of 74 and then we would be almost done with the result. 74 can be written as 492 which would end in 01 and so, we would get:
The unit’s place digit will be 1 while the tens place digit will be equal to the product of the tens place digit of the base (0) and the units place digit of the power (2) ie. 0*2=0
So, the last two digits are 01.
Find the last two digits of 223513
This becomes a bit longer than the previous one. First of all, we consider it as 23513 as the digits preceding 23 have no effect whatsoever on the last two digits (consequently, it would be important to note that the result would be the same even for 523513, 298723513 and so on)
Again, we split the original term into the form of a4k * (smaller term)
23512 * 23
(234)128 * 23
234 can be written as 292 which would have the same digits as 212=41
(41)128 * 23
41128 will have unit’s place digit = 1 and the tens place digit will be given by the product of the tens place digit of the base and the units place digit of the power ie (4*8=32) which would be 2. So, the last two digits would be 21.
Now we have to do 21*23 which will give us: 83 as the last two digits.
The base is even
This requires a different approach to the one we used earlier. We need to split the base into powers of 2 and powers of an odd number. The odd number we will deal with as was explained above. The power of 2 is what we are interested in. The things to note here are:
210 = 1024 ie. it ends in 24
220 ends in 76
230 ends in 24 and the sequence continues.
76*2n where n>1 will always end in the last two digits of 2n (say for example we get 25*76, we need not calculate. We can simply write the last two digits of 25 = 32 and we would be done). The trick lies in the expansion of the multiplication. See if you can figure it out.
Let’s solve a couple of examples to understand the process:
Find the last two digits of 10252
5152 * 252 is how we will split the number (remember that for any even number, as soon as we remove all the powers of 2, we are left with an odd number)
The last two digits of the first part of the number can be easily found to be 01.
The second part can be written as:
250 * 22
24*4 = 96
So, the last two digits would be 01*96 = 96
Find the last two digits of 362015
24030 * 34030
24*(34)1007 * 32
16 * (81)1007
16 * 61
which will be 76.
The base ends in 5
This is the simplest case and you can see that the cyclicity for the various numbers would depend on the tens place digit as can be seen:
|Last 2 digits of base||Cyclicity||Last 2 digits of base||Cyclicity|
|15||15, 25, 75, 25, 75||65||65, 25, 25|
|35||35, 25, 75, 25, 75||85||85, 25, 25|
|45||45, 25, 25||95||95, 25, 75, 25, 75|
|55||55, 25, 75, 25, 75||05||05, 25, 25|
As we have been seeing for the last few days now, when it comes to CAT, it is very important to understand the logic and method behind solving certain question-types. If someone does not know this concept, it would be very difficult to use either the cyclicity approach through observation for the last two digits or the remainder when divided by 100 approach. Historically, the questions have been straightforward and so, there should not be a reason to worry oneself by solving excessively large numbers.
All the best!