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Let no. of ₹100, ₹250 and ₹500 cheques are x, y and z respectively. And the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250.

So, __ x+y+z = 100__ &

By eliminating x from 2nd equation we get, 2(100-y-z)+5y+10z=305 => __3y+8z = 105__

The range of values of x, y and z are all non negative integers from 0 to 100. If we see equation

** 3y+8z = 105,** we will notice that for max value of z, the value of y should be minimum. So, we will hit and try the possible value of y starting from 0, for max value of z.

**z = (105 – 3y) / 8**

Here y = 0, 1, 2 will not result an integer value of z, but y = 3 will. **So, when y = 3, then z = 12.**

Answer is __12.__