Show that a positive integer, written in base 10, is divisible by 9 if the sum of its digits is divisible by 9.

The representation of the positive integer N in base p is denoted by \({(N)_p}\) .

If \({({5^{{{(126)}_7}}})_7} = {({a_n}{a_{n - 1}} \ldots {a_1}{a_0})_7}\) , find \({a_0}\) .

consider the decimal number \(A = {a_n}{a_{n - 1}} \ldots {a_0}\)     M1

\(A = {A_n} \times {10^n} + {a_{n - 1}} \times {10^{n - 1}} + \ldots + {a_1} \times 10 + {a_0}\)     M1

\( = {a_n} \times ({10^n} - 1) + {a_{n - 1}} \times ({10^{n - 1}} - 1) + \ldots + {a_1} \times (10 - 1) + {a_n} + {a_{n - 1}} + \ldots + {a_0}\)     M1A1

\( = {a_n} \times 99 \ldots 9(n{\text{ digits}}) + {a_{n - 1}} \times 99 \ldots 9(n - 1{\text{ digits}}) + \ldots + 9{a_1} + {a_n} + {a_{n - 1}} + \ldots + {a_0}\)     A1

all the numbers of the form 99…9 are divisible by 9 (to give 11…1),     R1

hence A is divisible by 9 if \(\sum\limits_{i = 0}^n {{a_i}} \) is divisible by 9     R1

Note: A method that uses the fact that \({10^t} \equiv 1(\bmod 9)\) is equally valid.

[7 marks]

by Fermat’s Little Theorem \({5^6} \equiv 1(\bmod 7)\)     M1A1

\({(126)_7} = {(49 + 14 + 6)_{10}} = {(69)_{10}}\)     M1A1

\({5^{{{(126)}_7}}} \equiv {5^{{{(11 \times 6 + 3)}_{10}}}} \equiv {5^{{{(3)}_{10}}}}(\bmod 7)\)     M1A1

\({5^{{{(3)}_{10}}}} = {(125)_{10}} = {(17 \times 7 + 6)_{10}} \equiv 6(\bmod 7)\)     M1A1

hence \({a_0} = 6\)     A1

[9 marks]

Questions similar to (a) have been asked in the past so it was surprising to see that solutions this time were generally disappointing.

In (b), most candidates changed the base 7 number 126 to the base 10 number 69. After that the expectation was that Fermat’s little theorem would be used to complete the solution but few candidates actually did that. Many were unable to proceed any further and others used a variety of methods, for example working modulo 7,

\({5^{69}} = {({5^2})^{34}}.5 = {4^{34}}.5 = {({4^2})^{17}}.5 = {2^{17}}.5\) etc

This is of course a valid method, but somewhat laborious.