If Tarun buys only pens costing Rs. 13 each or only pencils costing Rs. 5…

Key idea: If the amount leaves remainder 2 when divided by both 5 and 13, then the amount minus 2 is divisible by both 5 and 13, so it must be divisible by their lcm, 65. Thus valid amounts are numbers of the form 65k + 2.

• 457: 457 − 2 = 455, and 455 ÷ 65 = 7, so 457 = 65·7 + 2. This satisfies the condition.

• 782: 782 − 2 = 780, and 780 ÷ 65 = 12, so 782 = 65·12 + 2. This satisfies the condition.

• 577: 577 − 2 = 575. Although 575 is divisible by 5, 575 ÷ 13 = 44 remainder 3, so 575 is not divisible by 13 and hence not divisible by 65. Therefore 577 does not have the required remainder 2 on division by 13 and 5, so it cannot be the amount.

• 1042: 1042 − 2 = 1040, and 1040 ÷ 65 = 16, so 1042 = 65·16 + 2. This satisfies the condition.

Conclusion: 577 is the only number among the options that cannot be the amount because it does not leave remainder 2 when divided by 13.