To solve this problem, we need to break down the number into its place values. Let's denote the number as ABCD, where A is the thousands place, B is the hundreds place, C is the tens place, and D is the ones place. From the given information, we can form the following equations: B = A - 4, C = A + 1, and D = B + 5. By substituting the first equation into the second and third equations, we get C = (B + 4) + 1 = B + 5. Therefore, the number is 6953.
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Having: 9 ten-thousands → 90,000 the same number of thousands as ten thousands → 9 ten-thousands: 9 thousands → 9,000 8 fewer hundreds than ten-thousands → 9 ten-thousand: 9 - 8 = 1 hundreds → 100 6 more tens than hundreds → 1 hundreds: 1 + 6 = 7 tens → 70 6 fewer ones than thousands → 9 thousands: 9 - 6 = 3 ones → 3 → the number is 90,000 + 9,000 + 100 + 70 + 3 = 99,173.
7252
The number has 1 thousand. It has 8 more hundreds than thousands so it has 8 more hundreds than 1. That is, it has 9 hundreds. It has 1 fewer tens than hundreds so it has 1 fewer tens than 9. That is, it has 8 tens.It has 2 fewer ones than tens so it has 2 fewer ones than 8. That is, it has 6 ones. So the number is 1986.
5483
The number is 8,413