Units Digit of N

Since we are only interested in the units digit, we can work modulo 10. $N = 357^5 117^9 \equiv 7^5 7^9 = 7^{14} \pmod{10}$ But $7^4 = 2401 \equiv 1 \pmod{10}$ . Hence, $N \equiv 7^{14} = (7^4)^3 \times 7^2 \equiv 1^3 \times 49 \equiv 9 \pmod{10}$ .

N=(357)^5 * (117)^9

(357)^5 -- Ist part, (117)^9 -- 2nd part. In Ist part, for unit digit calculation it is sufficient to calculate unit digit of 7^5 i.e., 7. Similarly, in 2nd part, unit digit of 7^9 is 7.

Finally, the unit digit of N is the unit digit of the number that we get after multiplying the two calculated unit digits i.e., 9. ANSWER =9.

In this problem, we are only concerned with the units digit of N. Since the units digit of the product depends entirely upon the units digits of the factors, the units digit of $357^5 \cdot 117^9$ is the same as the units digit of $7^5 \cdot 7^9 = (7^{14})$ We can find the units digit using modulo 10. So, the simplified problem is: $7^{14} \equiv ? \pmod{10}$ Since it is very easy to find the square of a single-digit number, it is easy to find the square of any single-digit number in modulo 10. 7 squared, which is 49, would be equivalent to 9 because 9 is the units digit. We can evaluate the expression above by rewriting it as a product of squares, and simplifying by finding each square in mod 10: $7^{14}=(7^2)^7$ $7^2 \equiv 9 \pmod{10}$ $(7^2)^7 \equiv 9^7 \pmod {10}$ $9^7=(9^2)^3 \cdot 9$ $9^2 \equiv 1 \pmod {10}$ $(9^2)^3 \equiv 1 \pmod {10}$ $(9^2)^3 \cdot 9 \equiv 1 \cdot 9 \equiv 9 \pmod {10}$

Therefore, the units digit is 9.

357 \equiv 7 \pmod{10} 357^2 \equiv 357 \times 7 \equiv 2499 \equiv 9 \pmod{10} 357^4 \equiv 9 \times 9 \equiv 81 \equiv 1 \pmod{10} 357^5 \equiv 357 \times 1 \equiv 357 \equiv 7 \pmod{10}

117 \equiv 7 \pmod{10} 117^2 \equiv 117 \times 7 \equiv 819 \equiv 9 \pmod{10} 117^4 \equiv 9 \times 9 \equiv 81 \equiv 1 \pmod{10} 117^8 \equiv 1 \times 1 \equiv 1 \pmod{10} 117^9 \equiv 117 \times 1 \equiv 7 \pmod{10})

357^5 \times 117^9 \equiv 7 \times 7 \equiv 49 \equiv 9 \pmod{10} \text{Units digit of} 357^5 \times 117^9 \text{is} 9.

Consider only 7^5 and 7^9 because no.'s in ten's and hundred's place will have no effect on unit's place.So we observe a common pattern in successive multiplication of unit's place i.e. 7 at 7^1, 9 at 7^2, 3 at 7^3, 1 at 7^4 and 7 again at 7^5.We can infer that after every fourth power of 7 pattern of 7,9,3 and 1 at units place is repeated.So every no. whose unit's place's power can be written in the form 7^4n, where n is a whole no.. We can add the powers in multiplication so 7^5 and 7^9 = 7^14 =7^12+2=7^(4 3+2) = 7^(4 3) 7^2 So 7^(4 3) can be written in the form 7^4n = 1 (in its unit's place) and 7^2=9(in its units place. Therefore 7^14 = 1*9 = 9 (in its units place).

Given that you are only looking for the units digit, you can ignore the 350 in the 357 and the 110 in the 117, because the only thing that affects the units digit of the final product is the units digit of each digit

i.e. 357*357 = xxxxx9 etc.

Another way, therefore, of looking at it is that 357 * 357 (is congruent to) 7 * 7 [mod 10] which (is congruent to) 9 [mod 10]

Looking at the question, 7^5 (is congruent to) 7 [mod 10], and 7^9 is also congruent to 7 [mod 10] i.e. the answer is 7*7 [mod 10] and the units digit is 9

Considering that 357 = 3 \times 100 + 5 \times 10 + 7 \times 1 We can ignore the tenths and hundreds digits as when they are raised to any power as they will not have any effect on the final units digit of N. Thus units digit of N is simply determined by (7^5) \times (7^9)

The digits when raised to any power tend to have a pattern. Powers of 1 will simply always be 1 2: 2,4,8,6 3: 3,9,7,1 4: 4,6 5: 5 6: 6 7: 7,9,3,1 8: 8,4,2,6 9: 9,1 0: 0

Thus this problem simply reduces to finding the units digit of 7 \times 7 which gives 9.

We can very well do the problem using Congruent Modulo method.

in this particular question, without going for that lenghty method, try to use thi following method:

357 raise to some power will have the same number in its unit digit as 7 raised to that power.

Here, 357 is raised to power 5. => Unit digit will be sane as 7 raised to power 5 i.e.7 (as 7^5=16807)

For 117^9, do same method. You will get 7^9 's unit digit again as as 7.

Hence 7*7 =49. So unit digit is 9. Same will be the answer applying modulo arithmetic.

For the units digit, we only need to be concerned with the units's digit.

Do getting the last digit of 7^5 * 7^9 will do it for us.

Now simply prime factorize, and keep dropping the tenths place after every step. I will do 3 of them at a time, since 7^3 = 343

1) 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7 x 7

2) 3 x 3 x 3 x 3 x 7 x 7

3) 1 x 9

4) 9

And there is your answer :)

By modulo arithmetic, the last digit of a number, is equivalent to the number taken modulo 100. So, 357^5 \times 117^9 becomes 7^5 \times 7^9, which can further simplified to 7^14. Also, the last digit of a number raised to N = the last digit of the number raised to N (mod 4) So, last digit (unit's place) of 7^14 = last digit of 7^(14 mod 4) = last digit of 7^2 = last digit of 49 So the answer is 9.

Cyclicity is the rule by which we can find the unit digit of suc questions. According to it every single digit repeats its unit digit after every 4th power. 7^1 = 7 7^2 = 9 7^3 = 3 7^4 = 1 7^5 = 1 So we can observe that - 7^(4k+1) = 7 7^(4k+2) = 9 7^(4k+3) = 3 7^(4k) = 1 SO for the question 7^5 = 7 (unit digit) 7^9 = 7^(4 2+1) = 7 SO 7 7 - 9(unit digit)

Last digits in 7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4 = 1 Therefore the digits repeat after every fourth power. 7^5 = 7^(4+1)=7^1 = 7 7^9=7^(8+1) = 7 Therefore 7*7 = 9 Note: We are only talking about the last digit in the product

if i have an expression of the form (10a +b) * (10c +d) where the expression is actually a multiplication of two numbers like 39 54 where a=3,b=9,c=5,d=4 then the product will be 100ac+10(ad+bc)+bd; so the units place (without the 10s in multiplier is bd), to be accurate, its the modulo 10 of bd, but that doesnt affect the problem, as if you multiple 10 17 the units digit is modulo 10 of units digits of the numbers themselves, so question says (K+7)^5 multiplied by (V+7)^9 where K=350 & V=110; clearly the units digits are (7^5)%10 and (7^9)%10 and the products digit can be written as (7^14)%10; as its the property of (a b)%c that it may be expressed as [(a%c) (b%c)]%c;

so 7^14; if we raise 7 to 2, we get 49, ie. 9 in units place, 9 again multiplied by 7 is 63, 3 in units, 3 multiplied by 7 is 21, 1 in units, and again multiplied by 7 is 7, so multiplying by itself 4 completes one cycle, so we have 7(to begin with), we multiply 13 more times to get the 7^14 4 of em complete one cycle so 3 cycles and we are at 12 and we need one more and we are reset to 7. again multiplying we get 49, ie. units place 9

From the hint, the units digits of $N$ will be the value of modulo 10 of units digit of first factor $\times$ the units digit of second factor. To find the units digit of first factor, $(357)^5 \equiv (7)^5 \equiv (7)^4 \times 7 \equiv (49)^2 \times 7 \equiv (9)^2 \times 7 \\ \equiv (81)^2 \times 7 \equiv 7 \;(mod 10) \\$ Proceeding through similar steps for $(117)^9$ we get units digit for second factor as 7. Hence units digit of $N$ is $49 \;mod 10 = 9$

7 7=49 next 7 last digit of 49 =7*9=63 this process continous for 14 times

We seek the units digit, which is equivalent to working modulo 10. We can calculate that $7^4=2401\equiv 1 \pmod{10}$ . Thus, $N = 357^{5} 117^{9} \equiv 7^{5} \cdot 7 ^{9} \equiv 7^{14} \equiv 7^{2} \equiv 49 \equiv 9 \pmod{10}$ .

We can say that:

The 1st digit of that is: 7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

The cycle is repeating. Then,

Ther Last digit of 357^5 = 7

And the last digit of 117^9 = 7

Therefore,

The last digit of them is: 7*7 = 9

Final Answer:9

IOYIOYI