Question

Problem of the Week #4 1. An integer bis said to be divisible by an integer a 0, in symbols ab. if there exists some integer c such that b = ac. In other words, b is divisible by a if a goes into b with no remainder. For example, 30 is divisible by 5 (in symbols, 5 30 ) because 30 = 5 x 6. Problem of the Week: The following integers are all divisible by 31: 28272, 27683, 34441, 39618, and 55366. Use this fact to show that the following deter- minant 2 2 3 3 5 8 7 4 9 5 2 6 4 6 3 7 8 4 1 6 2 3 1 8 6 is divisible by 31 without directly evaluating the determinant. 2. Use your insights from the previous problem, to generate a 5 x 5 matrix such that the each entry of the matrix is an integer between 1 and 9, inclusive, and whose determinant is divisible by 541. Explain how you constructed the matrix to ensure that it would be divisible by 541.

Answer 1

Now in order to come up with 5*5 mcefox chose deteroom mot Should be divisible bey 541, we need to think of 5 meltiples of 541 etich are of 5 digits. 541 x 100 = 54100 542 X 50 = 7050 5 4 541 * 150 = 81 150 ez 5211 X 125 = 67625 8 i 15 541% 75 = 40575 no – on VNGGO Grooo 67 40 5

Ams: - Given, P8272, 27683, 34441, 39618 and 35 366 asu cell wińsible by 31. NOW, det A= 1 8 8 .7 2 2 7 6 8 3 3 4 4 4 1 3 9 6 1 8 5 5 3 6 6 Multiply GX10000 SXO", SX10?,C,XIO, i detA x 100 = 120104 8x10 &x10² 7410 2 2x104 4x103 6x10² 8x10 3 3x104 4x103 4x10² 4x10 ! 3x104 9x103 6x10² 1x10 581049 5x103 3x10² 6810 G GAG +5+Catç ...det A x 10 = 28272 - - - 27683 - 34441 39618 655366 Now, if we calculate the determinant treom this side, DO III let ax 10" - 28272x (a) + 27683 x (b) + 34441 X(c) + 39618 xD) + S5366 x Ca) Siosce all the numbers underlined are cinsible by 31. Therefore, deta is also divisibed by 31.