**Question:** We are interested to see divisibility checking of only prime numbers in base 10. For a prime P, you need to find the smallest positive integer N such that P’s divisibility testing is “N-sum” or “N-alterum”. Remember divisibility testing of 3 in base 10? It was simple, right? We need to add all the digits and then check if it is divisible by 3. Call this method “I-sum”. In case of testing of 11, we need to add all digits by alternating their signs. For example 1354379988104 = 11 *123125453464 and (4- 0+1-8+8-9+9-7+3-4+5-3+1) = O, which is divisible by 11 (0 = 0*11). Lees call this method “I-altersum”. In case of 7, we need to add all 3-digit-groups by alternating their signs. For 8618727529993 = 1231246789999 and (993- 529+727-618+8) = 581, which is divisible by 7 (581 = 7*83). Similarly, we call this method “3-altersum”. In similar Manner, 13’s checking is “3-altersum”. Develop a C++ program that solve the above problem, sample input output is listed below. Input: 1Output: Case 1: 77-altersum

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