Let x be the LCM of 32002– 1 and 32002 + 1. Find the last digit of x.
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3 Answers
206Since no one tried to solve this.....
31 = 3
32 = 9
33 = 27
34 = 81
35 = 243
Notice the last digits....from this we can infer that the last digits of powers of 3 would repeat after a cycle of 4 powers.
Now use concept of LCM to find x
206Clearly these two numbers would be consecutive even or odd numbers.
So, their lcm would simply be their products.
Now, we are only concerned with the last digits. So, lets find the last digit of the 2 nos concerned. As I have pointed out before, the last digit repeats itself after a cycle of 4 powers.
Thus, 2002 % 4 = 2
Hence, the last digit of 32002 would be the same as last digit of 32.
So, last digit = 9
____...__9 - 1 = __..._8
and
___....___9 + 1 = __...__0
So, Lcm = the product of the numbers,
hence, last digit of product = 8*0 = 0
Astha Gupta sir..wat does(%)...means...????Upvote·0· Reply ·2013-08-10 11:05:13- Sayantan Hazra it is the mod operator.....it gives you the remainder when a number is divided by another number
- Astha Gupta still didn't understand...can u explain it again....:(
- Astha Gupta sir..one more thing...why the no.s are consecutive even n odd no....i think the nos must be...3(2002)-1,3(2002),3(2002)+1....so they both will be either even or odd....
206When you divide a no by another no, you obtain two things:
1. Quotient
2. Remainder
When we do not need the quotient and simply need the remainder, we use the "mod" or the "modulo" operator(%). It divides the two given nos and returns only the remainder.
Eg: let us try to find remainder when 256, 287, 298 are divided by 10, say.
We know, the remainders will be 56, 87 and 98, respectively.
Mathematical representation:
256 mod 10 = 56
287 % 10 = 87
298 % 10 = 98
- Astha Gupta ok...i got it....thanku...:)