Web7 de abr. de 2024 · Solution For The difference between the highest and lowest five digits number using 0,3,6,8 and 9 digit (each digits using once time). [INV 2013] (1) 94941 (2) 61821 (3) 61740 (4) 67941 12. Web11 de abr. de 2024 · US bicycle and e-bike retail sales figures are hard to come by, and those that are published generally do not include the large number of bikes purchased online.It is highly likely that retail sales last year were well below the total of 13.4 million units imported and produced as large increases in bike inventory in warehouses have been …
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Web4 de nov. de 2024 · 5 Answers Sorted by: 2 You idea of using the modulus here is a good one: int input = in.nextInt (); int num = Math.abs (num); int highestDigit = -1; while (num > 0) { int digit = num % 10; if (digit > highestDigit) { highestDigit = digit; } num /= 10; } System.out.println ("Highest digit in " + input + " is: " + highestDigit); WebArrange the digits 2, 6, 0, and 1 so that you create the highest possible four-digit number. So the way I like to think about it is, if I'm trying to create as large of a number as possible, I want to put the largest numbers in the largest place value. imp foundry
How to find the largest digit of a given number - Stack Overflow
WebAnswer (1 of 6): The highest 4 hexadecimaldigit number is \text{FFFF}_{16}= (((15×16)+15)×16+15)×16+15= 16^4-1=2^{16}-1= \boxed{65535} Web25 de mar. de 2024 · Therefore, the greatest 5 digit number using the digits 1, 2, 5 is 55521. To get the smallest number, the smallest digit 1 is placed at ten thousands-place, next greater digit 1 at thousand’s place, still greater digit 1 at hundred’s place, next greatest digit 2 is placed at ten’s place, and greatest digit 1 at one’s or units place. WebYour structure is basically right - you get a number from the user in main, pass it to maximum as a parameter, maximum returns the answer to main, main prints it. You just got your variables all mixed up in the maximum method. 0 0 hfx642 11 Years Ago Think about what you are trying to accomplish in your maximum () method. impfort