1234, 1345, 1456, 1567, 1678, 1789, 2345, 3456 4567 5678 6789 0000 9999 8888 7777 6666 5555 4444 3333 2222 1111 1212 1313 1414 1515 1616 1717 1818 1919 1010 2121 2323 2424 2525 2626 2727 2828 2929 3030 3131 3232 3434 3535 3636 3737 3838 3939 4040 4141 4242 4343 4545 4646 4747 4848 4949 5050 5151 5252 5353 5454 5656 5757 5858 5959 6060 6161 6262 6363 6464 6565 6767 6868 6969 7070 7171 7272 7373 7474 7575 7676 7878 7979 8080 8181 8282 8383 8484 8585 8686 8787 8989 9090 9898 9797 9696 9191 9292 9393 9494 9595 1231 1241 1251 1261 1271 1281 1291 1222 1232 1242 1252 1262 1272 1282 1292 1213 1214 1215 1216 1217 1218 1219 121 these are a few but there are millions
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To find the smallest odd number that can be made using the numbers 4, 5, 3, and 6, we need to consider all possible combinations. The odd numbers that can be formed are 3 and 5. The smallest odd number would be 3, which can be formed using the numbers 3 and 6 by ignoring the digit 6.
0 through 9 is 10 digits. For the first number you have 10 choices, same for the second and third and forth, so your total amount of possible combinations (assuming you can repeat) is: 10 * 10 * 10 * 10 = 104 = 10,000 (ten thousand) If repetition is not allowed, then for the first you have 10 choices, second you have 9 (every digit except the one you used for the first), third you have 8 and forth you have 7, giving you: 10 * 9 * 8 * 7 = 5040
The answer is 15.
4x4x3+2
To find the numbers that add up to 536, we need to consider all possible combinations. One way to approach this is by using a systematic method like trial and error or algebraic equations. For example, we could set up an equation like x + y = 536 and solve for x and y. Another approach could be to break down 536 into its prime factors and then find combinations that add up to the target number. Ultimately, there are multiple ways to find the numbers that add up to 536, depending on the context and constraints of the problem.