The largest 5-digit number having three different digits is
99987
The question asks for the largest possible number that has exactly five digits and is formed using only three different digits. A 5-digit number is any integer from 10000 to 99999.
The key constraint is that we can only use three unique digits out of the possible ten digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). To create the Largest 5-Digit Number, we should choose the largest possible digits available and place them in the highest value positions.
To make a number as large as possible, we should use the largest digits. The largest digits are 9, 8, 7, 6, and so on. Since we are limited to using only three different digits, we should pick the three largest digits available. These are 9, 8, and 7.
A 5-digit number has positions for ten thousands, thousands, hundreds, tens, and units. To make the number largest, the digit in the ten-thousands place must be the largest possible. We should use the largest digit from our chosen set {9, 8, 7}, which is 9.
So, the number starts with 9xxxx.
Next, the thousands place should also be as large as possible. Using 9 again makes the number larger: 99xxx.
Similarly, the hundreds place should be 9 to maximize the value: 999xx.
We now have the number 999xx. We have used the digit 9 three times. We still need to fill the tens and units places, and we must ensure the final number uses exactly three different digits. Our chosen digits are 9, 8, and 7. We have used 9. We must now introduce 8 and 7 into the remaining places.
To make the number largest, the next largest digit from our set {8, 7} should go into the tens place. This is 8. So the number becomes 9998x.
Finally, the remaining digit from our set {7} must go into the units place. This is 7. So the number becomes 99987.
Let's check if 99987 meets the criteria:
Therefore, 99987 is the Largest 5-Digit Number having three different digits.
If m and n are two positive real numbers such that 9m2 + n2 = 40 and mn = 4, then the value of 3m + n is:
Which composite number can divide the sum of the first 12 natural numbers?
Consider the following statements :
1. If n is a natural number, then the number \(\frac{n\left(n^2+2\right)}{3}\) is also a natural number.
2. If m is an odd integer, then the number \(\frac{\mathrm{m}^4+4 \mathrm{~m}^2+11}{16}\) is an integer.
Which of the statements given above is/are correct ?
If 1 is added to the greatest 7-digit number, it will be equal to
The product of successor and predecessor of 999 is