If lucky no. is defined as the no. whose sum of digits is 7, then lucky nos. are in sequence: 7,16,25,34... If 7=A1, 16=A2, 25=A3 and so on; find A65 and A325. (After adding digits once, you cannot add them more times for example:583 is not a lucky no. as you have to add its digits twice to get 7)

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For single digit: 7

For two digits: we have \( \overline{a_1 a_2} \), where \(a_1 + a_2=7 \), \( 0 < a_1 \leq 7 \), \(0 \leq a_2 < 7 \) which gives \(16,25,34,43,52,61,70 \).

For three digits: we have \( \overline{a_1 a_2 a_3} \), where \(a_1 + a_2 + a_3 =7 \), \( 0 < a_1 \leq 7 \), \(0 \leq a_2, a_3 < 7 \).

For four digits: we have \( \overline{a_1 a_2 a_3 a_4} \), where \(a_1 + a_2 + a_3 +a _4=7 \), \( 0 < a_1 \leq 7 \), \(0 \leq a_2, a_3, a_4 < 7 \).

For five digits: we have \( \overline{a_1 a_2 a_3 a_4 a_5} \), where \(a_1 + a_2 + a_3 +a _4 +a_5 =7 \), \( 0 < a_1 \leq 7 \), \(0 \leq a_2, a_3, a_4, a_5 < 7 \).

Since \(1+7 + 28 < 65 < 1+7 + 28 + 84 \),

\(A_{65} \) must be a four digit number. \(65-1-7-28 = 29\).

\( 29 - { 6+2 \choose 2 } = 1 \). So \( A_{65} \) must the smallest four digit number with \(a_1 = 2 \).

\(\Rightarrow \boxed{ A_{65} = 2005 } \)

Similarly \(1+7 + 28 + 84 < 325 < 1+7 + 28 + 84 + 210 \),

\(A_{325} \) must be a five digit number. \(325-1-7-28-84 = 205\). Or \(325-1-7-28-84 -210 = -5\)

\( \Rightarrow A_{330} = 70000 \)

\( \Rightarrow A_{329} = 61000 \)

\( \Rightarrow A_{328} = 60100 \)

\( \Rightarrow A_{327} = 60010 \)

\( \Rightarrow A_{326} = 60001 \)

\( \Rightarrow \boxed{ A_{325} = 52000 } \) – Pi Han Goh · 3 years, 8 months ago

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