Nice Maths Question...

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)

Note by Partho Paul
7 years, 9 months ago

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Comments

Stars and bars is applicable here.

For single digit: 7

(1 of them)

For two digits: we have a1a2 \overline{a_1 a_2} , where a1+a2=7a_1 + a_2=7 , 0<a17 0 < a_1 \leq 7 , 0a2<70 \leq a_2 < 7 which gives 16,25,34,43,52,61,7016,25,34,43,52,61,70 .

(7 of them)

For three digits: we have a1a2a3 \overline{a_1 a_2 a_3} , where a1+a2+a3=7a_1 + a_2 + a_3 =7 , 0<a17 0 < a_1 \leq 7 , 0a2,a3<70 \leq a_2, a_3 < 7 .

  • When a1=1a_1 = 1 , there's (6+11) { 6+1 \choose 1 } solution

  • When a2=2a_2 = 2 , there's (5+11) { 5+1 \choose 1 } solution

  • ...

  • when a1=7a_1 = 7 , there's (0+11) { 0+1 \choose 1 } solution.

(28 of them)

For four digits: we have a1a2a3a4 \overline{a_1 a_2 a_3 a_4} , where a1+a2+a3+a4=7a_1 + a_2 + a_3 +a _4=7 , 0<a17 0 < a_1 \leq 7 , 0a2,a3,a4<70 \leq a_2, a_3, a_4 < 7 .

  • When a1=1a_1 = 1 , there's (6+22) { 6+2 \choose 2 } solution

  • When a1=2a_1 = 2 , there's (5+22) { 5+2 \choose 2 } solution

  • ...

  • When a1=7a_1 = 7 , there's (0+22) { 0+2 \choose 2 } solution

(84 of them)

For five digits: we have a1a2a3a4a5 \overline{a_1 a_2 a_3 a_4 a_5} , where a1+a2+a3+a4+a5=7a_1 + a_2 + a_3 +a _4 +a_5 =7 , 0<a17 0 < a_1 \leq 7 , 0a2,a3,a4,a5<70 \leq a_2, a_3, a_4, a_5 < 7 .

  • When a1=1a_1 = 1 , there's (6+33) { 6+3 \choose 3 } solution

  • When a1=2a_1 = 2 , there's (5+33) { 5+3 \choose 3 } solution

  • ...

  • When a1=7a_1 = 7 , there's (0+33) { 0+3 \choose 3 } solution

(210 of them)

Since 1+7+28<65<1+7+28+841+7 + 28 < 65 < 1+7 + 28 + 84 ,

A65A_{65} must be a four digit number. 651728=2965-1-7-28 = 29.

29(6+22)=1 29 - { 6+2 \choose 2 } = 1 . So A65 A_{65} must the smallest four digit number with a1=2a_1 = 2 .

A65=2005\Rightarrow \boxed{ A_{65} = 2005 }

Similarly 1+7+28+84<325<1+7+28+84+2101+7 + 28 + 84 < 325 < 1+7 + 28 + 84 + 210 ,

A325A_{325} must be a five digit number. 325172884=205325-1-7-28-84 = 205. Or 325172884210=5325-1-7-28-84 -210 = -5

A330=70000 \Rightarrow A_{330} = 70000

A329=61000 \Rightarrow A_{329} = 61000

A328=60100 \Rightarrow A_{328} = 60100

A327=60010 \Rightarrow A_{327} = 60010

A326=60001 \Rightarrow A_{326} = 60001

A325=52000 \Rightarrow \boxed{ A_{325} = 52000 }

Pi Han Goh - 7 years, 9 months ago
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