Question

find the unit digit in the sum of 124^372+124^373​

Answer 1

Answer:

0

Step-by-step explanation:

There's a concept of cyclic digits

for example, if a number with unit digit = 1 is raised to a natural number than

the unit digit of number would still remain one for example

$1^{14} = 1\\\\11^{2} = 121\\\\11^{3} = 1331\\\\11^{4} = 14641$

Now if we see for numbers having unit digit 2 Then,

$2^{1} = 2\\\\2^{2} = 4\\\\2^{3} = 8\\\\2^{4} = 16$

You can generalize this for any number with unit digit 2 for example 12,22,222 and 562 etc.

Now You can also see that

$2^{4n}, 2^{4n+1},2^{4n+2},2^{4n+3}$ would respectively end in unit digits 6,2,4,8

Similarly

For 3 : $3^{4n},3^{4n+1},3^{4n+2},3^{4n+3}$ The unit digits are 1,3,9,7

For 4 : $4^{2n},4^{2n+1}$ end in 6 and 4 respectively

For 5 : $5^{n}$ will always end in 5

For 6 : $6^{n}$ will always end in 6

For 7 : $7^{4n},7^{4n+1},7^{4n+2},7^{4n+3}$ The unit digits are 1,7,9,3 respectively

For 8 : $8^{4n},8^{4n+1},8^{4n+2},8^{4n+3}$ The unit digits are 6,8,4,2

For 9 : $9^{2n},9^{2n+1}$ The unit digits are 1 and 9 respectively.

For 10 : $10^{n}$ will always end in zero

Now you can just simply use this information to find the answer.

I have mentioned all the digits so that you may use them for future reference.

$124^{372} + 124^{373} \\\\124^{2\times196} + 124^{2\times196+1}\\\\4^{2n}+4^{2n+1}\\\\6+4\\\\0$

Hence 0 is the correct unit digit for the following expression.

Might Help! Thanks!