1.EIf an odd number is written as the sum of two prime numbers then one of those primes is2, since 2 is the only even prime. However, 9 is not prime so 11 cannot be written as thesum of two primes. Note that 5 = 2 + 3; 7 = 2 + 5; 9 = 2 + 7; 10 = 3 + 7,so 11 is the onlyalternative which is not the sum of two primes.2.BThe interior angles of an equilateral triangle, square, regular pentagon are,,respectively.The sum of the angles at a point is.So.60°90°108°360°θ=360− (60+90+108) =1023.DThe cost now is.(70+4×5+6×2)=£1.02p4.BOne hundred thousand million is. So the number of stars is.102×103×106=10111011×1011=10225.CLet the required addition be, where,,,,,are single, distinct digits.To make this sum as large as possible, we need,,(the tens digits) as large as possible;sothey must be 7, 8, 9 in some order.Then we need,,as large as possible, so 4, 5, 6 insome order. Hence the largest sum is.‘ab’+‘cd’+‘ef’a b c d e fa c eb d f10(7+8+9) + (4+5+6) =10×24+15=2556.CIn order to be a multiple of 15, a number must be a multiple both of 3 and of 5. So itsunits digit must be 0 or 5.However, the units digit must also equal the thousands digitand this cannot be 0, so the required number is of the form. The largest such four-digit numbers are 5995, 5885, 5775. Theirdigit sums are 28, 26, 24 respectively. In orderto be a multiple of 3, the digit sum of a number must also be a multiple of 3, so 5775 isthe required number. The sum of its digits is 24.‘5aa5’7.DAdd the first and third equations:, so.Add the first two equations:, so. Substitute forandin the first equation:so.Therefore.