####
**MATHS TRICK-18**

#### This is useful for those preparing for different Mathematical Entrance exams like CSIR-NET, TIFR, ISI, CMI Entrances. Here it is:

**Cycle in the Last Digit**

The last digit — the ones place — of a decimal integer d is the
remainder of the division d/10. Equivalently, the last digit is the result of
the operation

*d*mod*10*, when following the convention that the least non-negative value — the common residue — is returned. Modular arithmetic, combined with iterative generation of the positive powers of two, allows us to show the cycle in the last digit:
We start with 2, compute it mod 10, multiply that result by 2,
compute

*it*mod 10, etc. From this it’s clear the pattern will repeat, once a previous result — 2 in this case, at step 5 — is obtained. This shows that the numbers 2^{n}, n ≥ 1, cycle through the four ending digits 2, 4, 8, and 6.
The cycle implies that powers of two with the same ending digit
are related, their exponents differing by a multiple of four:

**Ends in 2:**2^{1}, 2^{5}, 2^{9}, 2^{13}, 2^{17}, … .**Ends in 4:**2^{2}, 2^{6}, 2^{10}, 2^{14}, 2^{18}, … .**Ends in 8:**2^{3}, 2^{7}, 2^{11}, 2^{15}, 2^{19}, … .**Ends in 6:**2^{4}, 2^{8}, 2^{12}, 2^{16}, 2^{20}, … .

You can express these relationships more succinctly using
the laws of exponents, showing explicitly that the
ending digit of the first four positive powers of two determine the ending
digit of

*all*positive powers of two:**Ends in 2:**2^{1}·2^{4k}, or 2^{1+4k}, k ≥ 0.**Ends in 4:**2^{2}·2^{4k}, or 2^{2+4k}, k ≥ 0.**Ends in 8:**2^{3}·2^{4k}, or 2^{3+4k}, k ≥ 0.**Ends in 6:**2^{4}·2^{4k}, or 2^{4+4k}, k ≥ 0.

**Problem : [TIFR-2010] The last digit of 2**

^{80 }is**(a)**

**2 (b) 4 (c) 6 (d) 8**

^{}

Solution : (c) Apply
this trick here:

**Ends in 6:**2

^{4}·2

^{4k}, or 2

^{4+4k}, k ≥ 0. Put k=19.