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How many multiples of 3 are there from 3 to 123?
3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 99 102 105 108 111 114 117 120 123
How to judge whether a number is a multiple of 123?
456000 ÷ 123 = 3707.32
be
3708 × 123 = 456084
Fill in 084 in brackets
123.234.345.456. Why are all such numbers multiples of 3
Because the sum of all the numbers is a multiple of three
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- 1. Python looks for all the integer multiples of another number in a range, and calculates how many multiples there are This is a problem. I really can't do it. I use Python 2.7. I want to use for loop 1) Create program count_ Multiple () which takes three non negative integers: base, start and stop, prints each integer multiple of base which occurs between start and stop (including start but not including stop) on a separate line, And returns the number of multiples found. If base = 3, there are two integer multiples between start = 9 and stop = 15, 9 and 12, but excluding 15. The east way to test while one number is an integer multiple of another is with the% operator \x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05 \x05\x05\x05\x05 \x05\x05\x05 \x05\x05 \x05\x05\x05 \x05\x05\x05\x05 \x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05 2）.Write a function user_ input_ multiples() which takes a single integer input base.This function will get start and stop values from the user with two calls to raw_ input(),call count_ multiples() to determine the number of integer multiples of base between the user specified start and stop,and then ask again for new start and stop values.The function will continue asking for new start and stop values until at least one of the following cases occurs: \x05\x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05\x05\x05 The user enters a negative value for start or stop. \x05\x05\x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05\x05\x05 The user enters a value for stop which is less than the value for start. \x05\x05\x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05\x05 \x05\x05\x05\x05\x05\x05\x05\x05\x05 The function count_ multiples() returns zero (eg:there were no multiples between start and stop). Once the function stops asking for input,it will return the total number of multiples found (the total over all calls to count_ multiples()). Hint:You will want to use a while loop for this function. English is a little too much. I'm a little annoyed. Please forgive me If you don't have time, give me a detailed idea or direction
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