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A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, even. Find the number.

Solution:

Given, a perfect square number has four digits, none of which is zero.

The digits from left to right have values that are: even, even, odd, even.

We have to find the number.

Consider abcd is a perfect square

where, a = even number

b = even number

c = odd number

d = even number

We know the square root of the smallest four digit number is 32 × 32 = 1024

We know that the square root of the largest four digit number is 99 × 99 = 9801

This implies that the number lies between 32 and 99.

Since the unit place number is even, abcd will be the square of an even number.

The value of the digit a has to be 2, 4, 6, 8.

It is clear that the squares will be in the range 42 to 98 as 32, 34 etc do not satisfy the given conditions.

Now, 94 × 94 = 8836

Here, a = 8; b = 8; c = 3 and d = 6

Therefore, the required number is 8836.

✦ Try This: A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, odd, even, odd. Find the number.

☛ Also Check: NCERT Solutions for Class 8 Maths

NCERT Exemplar Class 8 Maths Chapter 3 Problem 138

A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, even. Find the number

Summary:

A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even, odd, even. The number is 8836.

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