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question:What is the smallest positive integer that satisfies the congruence 4x equiv 17 pmod{31}?
answer:The equation given is 4x equiv 17 pmod{31}. First, notice that multiplying 4 by 8 results in 32, which is congruent to 1 modulo 31, simplifying the equation. We multiply both sides of the initial congruence by 8: [ 8 cdot 4x equiv 8 cdot 17 pmod{31}. ] This simplifies to: [ 32x equiv 136 pmod{31}. ] Since 32 equiv 1 pmod{31}, we have: [ x equiv 136 pmod{31}. ] To simplify 136 pmod{31}, we compute 136 div 31 = 4 remainder 12, thus: [ x equiv 12 pmod{31}. ] Thus, boxed{12} is the smallest positive integer that satisfies the congruence.
question:f a = 3 and b = 4, then what is the value of frac{a^3 + 3b^2}{9}?
answer:Given a = 3 and b = 4, we need to calculate frac{a^3 + 3b^2}{9}. 1. Calculate a^3 = 3^3 = 27. 2. Calculate b^2 = 4^2 = 16, then 3b^2 = 3 times 16 = 48. 3. Sum the results: 27 + 48 = 75. 4. Divide by 9: frac{75}{9} = boxed{8.333ldots} or boxed{frac{25}{3}}.
question:w many ordered pairs (m,n) of positive integers, with m ge n, have the property that their squares differ by 120? A) 3 B) 4 C) 5 D) 6 E) 7
answer:To find the number of ordered pairs (m,n) where m ge n and m^2 - n^2 = 120, we proceed in a similar fashion to the given template: [ m^2 - n^2 = (m+n)(m-n) = 120. ] List the factor pairs of 120 where the first element is less than or equal to the second (since m+n must be greater or equal to m-n): [ (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12). ] For each pair (a, b) where a leq b, solve for m and n: [ m + n = b, quad m - n = a. ] [ m = frac{b + a}{2}, quad n = frac{b - a}{2}. ] (begin{align*} 1. & (1, 120): m = frac{121}{2}, n = frac{119}{2} text{ (not integers)} 2. & (2, 60): m = frac{62}{2} = 31, n = frac{58}{2} = 29 text{ (integers)} 3. & (3, 40): m = frac{43}{2}, n = frac{37}{2} text{ (not integers)} 4. & (4, 30): m = frac{34}{2} = 17, n = frac{26}{2} = 13 text{ (integers)} 5. & (5, 24): m = frac{29}{2}, n = frac{19}{2} text{ (not integers)} 6. & (6, 20): m = frac{26}{2} = 13, n = frac{14}{2} = 7 text{ (integers)} 7. & (8, 15): m = frac{23}{2}, n = frac{7}{2} text{ (not integers)} 8. & (10, 12): m = frac{22}{2} = 11, n = frac{2}{2} = 1 text{ (integers)} end{align*}) Valid pairs: (31, 29), (17, 13), (13, 7), (11, 1). Each yields exactly one valid ordered pair (m, n) since m geq n. There are 4 valid ordered pairs (m, n) where m^2 - n^2 = 120. boxed{The correct answer is B) 4.}
question:Starting with a given positive integer ( n_{0} > 1 ), players A and B take turns selecting numbers ( n_{1}, n_{2}, cdots ) according to the following rules: When ( n_{2k} ) is selected, player A can choose any integer ( n_{2k+1} ) such that ( n_{2k} leqslant n_{2k+1} leqslant n_{2k}^{2} ). When ( n_{2k+1} ) is selected, player B can choose ( n_{2k+2} ), where ( n_{2k+2} ) is an integer such that ( frac{n_{2k+1}}{n_{2k+2}} ) is a positive integer power of a prime number. Player A wins if they can first select 1990, and player B wins if they can first select 1. Determine: 1. For which ( n_{0} ) does player A have a winning strategy? 2. For which ( n_{0} ) does player B have a winning strategy? 3. For which ( n_{0} ) do both players lack a guaranteed winning strategy? (From the 31st IMO)
answer:Problem Analysis: Given a strictly positive integer ( n_0 > 1 ), two players, ( A ) and ( B ), take turns choosing integers ( n_1, n_2, cdots ) according to the following rules: 1. ( A ) chooses ( n_{2k+1} ) such that ( n_{2k} leq n_{2k+1} leq n_{2k}^2 ). 2. ( B ) chooses ( n_{2k+2} ) such that the ratio ( frac{n_{2k+1}}{n_{2k+2}} ) is a prime integer. - ( A ) wins if they can reduce ( n ) to 1990. - ( B ) wins if they can reduce ( n ) to 1. The goal is to determine the conditions under which each player has a winning strategy. Part 1: ( A )'s Winning Strategy # Inductive Approach: 1. **Starting Induction Hypothesis**: Assume ( W ) is the set of integers for which ( A ) has a winning strategy. We know the smallest ( n ) such that: [ {1990, 1989, cdots, 45 } subset W ] 2. **Inductive Step**: - For ( 1 < n_0 < 45 ), need to demonstrate it belongs to ( W ) given the properties and rules defined. # Establishing The Winning Set: - For some ( m ), if ( {m, m+1, ..., 1990} subset W ), show integers satisfying ( S leq 1990 ): [ S = p^kq^r (p, q text{ primes}) ] such that ( frac{S}{p^r} geq m ): - Any integer with ( n_0 ) such that ( sqrt{S} leq n_0 < m ), is in ( W ). Example Cases: - **For ( m = 45 )**: Choose ( S = 2^2 times 3 times 5 times 7 = 420 ), [ sqrt{420} < 21 < 45 ] Therefore ( {21, 22, cdots, 44 } subset W ) - **For ( m = 21 )**: Choose ( S = 2^3 times 3 times 7 = 168 ), [ sqrt{168} < 13 < 21 ] Therefore ( {13, 14, ..., 20 } subset W ) - **For ( m = 13 )**: Choose ( S = 3 times 5 times 7 = 105 ), [ sqrt{105} < 11 ] Therefore ( {11, 12 } subset W ) - **For ( m = 11 )**: Choose ( S = 2^2 times 3 times 5 = 60 ), [ sqrt{60} < 8 < 11 ] Therefore ( {8, 9, 10 } subset W ) Thus, [ {8, 9, 10, ..., 1990 } subset W ] # For ( n_0 > 1990 ): Choose ( r ) such that ( 2^r cdot 3^2 < n_0 leq 2^{r+1} cdot 3^2 )< n^{2} ). - Choose ( n_1 = 2^{r+1} cdot 3^2 ), whatever ( B ) chooses for ( n_2 ), ( 9 leq n_2 < n_0 ). By finite steps eventually ( A ) ensures a winning set ( W ). Conclusion: Since ( {8, 9, 10, cdots }subset W), [ boxed{ {8, 9, 10, cdots} subseteq W }] Part 2: ( B )'s Winning Strategy For ( n_0 leq 5 ): [ n_0 in {2, 3, 4, 5} ] - **Reasoning**: - Because the smallest product of three distinct primes exceeds ( 5^2 ), ( A ) cannot form an integer such that ( frac{n}{q^r} leq 1 ). Therefore, if ( n_0 leq 5 ), ( B) always wins through the above mechanism. [ boxed{ n_0 in {2, 3, 4, 5} } ] Part 3: Neither Player Has a Forced Win For ( n_0 = 6 ) or ( n_0 = 7 ): - **Reasoning**: - ( A ) must choose a number such that ( n_2k+3 = 5 ), thus forcing (B to eventually reach 1. - Combinations are finite and do not guarantee a certain win. [ boxed{ n_0 in {6, 7} }] # Conclusion: - (A) wins if ( n_0 geq 8). - (B) wins if ( n_0 leq 5). - Neither wins certainly if ( n_0 = 6) or ( 7 ).