Find the last digit of 3278 123
WebMay 21, 2024 · To find : The unit digit of the expression? Solution : First we determine the cyclicity of number 9. The cyclicity of 9 is 2. Now with the cyclicity number i.e. with 2 divide the given power i.e. 85 ÷ 2 The remainder will be 1. The required answer is 9 raised to the power 1 is 9. Therefore, The unit digit of is 9. Advertisement WebOct 12, 2024 · Print. In ( (36472)^123!) ,the last two digits of 123! would be 00 as it is a factorial and hence we can say that it is divisible by 4.The unit digit depends on the unit …
Find the last digit of 3278 123
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Webfind the unit digit of 3278^1237 - YouTube. #RevoClasses #ShortTrickMaths #Aptitudefind the unit digit of 3278^1237Join the Telegram Channel by clicking the link … WebIn previous videos, we've already talked about the idea of place value, and a place value table or a place value chart is just a way to say how much we have, how much value we have in each place in a very, very clear way. So, here they say use the place value chart to write 60,229. And, this is the place value table or place value chart right ...
WebJul 30, 2014 · You already have the mechanism to extract the last digit from a number. You just need to extend this by using use division to "shift" the digits the required number of places. For example, if you have the number 1234, to get the second digit you can divide by 10, to get 123. Using 123 mod 10 will give you the digit 3. WebFeb 10, 2024 · The position of the last significant number is indicated by underlining it. For multiplication and division operations, the result should have no more significant figures than the number in the operation with the least number of significant figures.
WebNov 25, 2008 · That works. I would have just said since 7^400=1 mod 1000, then 7^10000=1 mod 1000. So if you let x=7^9999. Then you want to solve 7*x=1 mod 1000. … WebFind unit digit in the product : (6374)1793 x (625)317 x (341)491 Solution : In (6374)1793, unit digit is 4. The cyclicity of 4 is 2. Dividing 1793 by 4, we get 1 as remainder. 41 = 4 So, the unit digit of (6374)1793 is 4. In (625)317, unit digit is 5. Since 5 has the cyclicity 1, the unit digit of (625)317 is 5. In (341)491, unit digit is 1.
WebFeb 19, 2024 · In ( (36472)^123!) ,the last two digits of 123! would be 00 as it is a factorial and hence we can say that it is divisible by 4.The unit digit depends on the unit digit of …
WebNote that φ ( 100) = 100 ( 1 − 1 2) ( 1 − 1 5) = 40 using Euler's product formula, so since 3 and 100 are coprime, 3 φ ( 100) ≡ 3 40 ≡ 1 (mod 10) and so 3 885 ≡ ( 3 40) 22 ⋅ 3 5 ≡ 1 22 ⋅ 3 5 ≡ 3 5 ≡ 243 ≡ 43 (mod 100), i.e. the last two digit of 3 885 are 43. Edit: I misread the question as asking only about the last digit of 3 885. Share Cite class 10 cbse maths pdfWebThe last digit of 2345714 is 4 because 2345714 = 234571*10 + 4. The last 3 digits of 2345714 are 714 because 2345714 = 2345*1000 + 714 and so on. More to the point, ... download free unreal engine assetsWeb1 day ago · Find many great new & used options and get the best deals for The Taking of Pelham 123 (Blu-ray+Digital Copy,Canadian) Action - Free Shipping at the best online prices at eBay! Free shipping for many products! download free undertaleWebLong Multiplication Example: Multiply 234 by 56. Long Multiplication Steps: Stack the numbers with the larger number on top. Align the numbers by place value columns. … download free unlimited vpnWebThe two last digits of the number 9^123, therefore, is not difficult to calculate : they are 29. Therefore, the last two digits of the number 3^123 + 7^123 + 9^123 you can easily find … class 10 cbse maths tb pdfWebn = 56789 lastdigit = int (repr (n) [-1]) # > 9 Convert n into a string, accessing last element then use int constructor to convert back into integer. For a Floating point number: n = 179.123 fstr = repr (n) signif_digits, fract_digits = fstr.split ('.') # > ['179', '123'] signif_lastdigit = int (signif_digits [-1]) # > 9 Share Improve this answer class 10 cbse maths exercise 5.1WebThe answer is [math]60 [/math]. It can be solved using congruences modulo [math]100 [/math] but there’s an easier way. I can divide [math]234 [/math] by [math]2 [/math] and [math]345 [/math] by [math]5 [/math] to make an equivalent product as follows: [math]\quad 123\times 234\times 345\times 456\times 567\times 678\times 789 [/math] download free unlocker