If 2ab09 is a perfect square, what is the maximum value of (10a + 3b)?

To solve this problem, we need to find out which digits 'a' and 'b' can take such that the number 2ab09 is a perfect square. After identifying suitable values for 'a' and 'b', we aim to maximize the expression 10a + 3b.
A number is a perfect square if it is the square of an integer. Therefore, 2ab09 should be equal to n² for some integer n.
Let's break down the number:

It is a 5-digit number with 2 at the ten-thousands place, "a" at the thousands place, "b" at the hundreds place, and 09 at the tens and units place.
Therefore, the number can be expressed as: 20000 + 1000a + 100b + 9 .

Since it ends in 9, the perfect square must be the square of a number ending in 3 or 7 (because 3² = 9 and 7² = 49).
Next, we will calculate possible n-values by taking the square root range:

For a 5-digit number starting with 2, the range for n is approximately between 141 (since 141² = 19881) and 158 (since 158² = 24964) because any higher squares would exceed a 5-digit number starting with 2.

To find values for 'a' and 'b', try calculating n² where n is from 141 to 158 to find one that matches the form 2ab09 exactly:

For example, if n = 147, then n² = 147² = 21609, matching the form 2ab09 with a = 1, b = 6. Another test could be n = 153, then n² = 23409, giving a = 3, b = 4.

Now, calculate 10a + 3b for the potential solutions:

For (a, b) = (1, 6), 10a + 3b = 10 * 1 + 3 * 6 = 28
For (a, b) = (3, 4), 10a + 3b = 10 * 3 + 3 * 4 = 42

Among these, the maximum value is 42 for (a, b) = (3, 4).
Therefore, the maximum value of 10a + 3b is 42 when a = 3 and b = 4.

Answered by IsabellaRoseDavis | 2025-07-22

The maximum value of 10 a + 3 b for the number 2ab09 being a perfect square is 42, achieved when a = 3 and b = 4.
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Answered by IsabellaRoseDavis | 2025-07-23